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RD Sharma Solutions Class 12 Mathematics Chapter 21 RE

RD Sharma Solutions Class 12 Mathematics Chapter 21 RE

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 05:45 PM IST

RD Sharma Solutions for Class 12 are made to help students perform well and learn various concepts. RD Sharma books have a good reputation for being informative and comprehensive. Class 12 Maths is a challenging subject that requires a lot of practice. RD Sharma Class 12th Exercise RE material helps students practice better as it contains solved examples that have step-by-step answers. This is an excellent alternative for preparation as it follows the CBSE syllabus and is prepared by subject experts. RD Sharma Class 12 Chapter 21 Exercise RE also helps students with Revision as the solutions are easy to understand and are available in one place.

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter21 RE Differential Equation- Other Exercise

Differential Equations Excercise:RE

Differential Equations Exercise Revision Exercise Question 1(i)

Answer:
Order2, degree1
Hint:
Check the highest order of derivate to find the order and check the power of derivative to find degree.
Given:
(dsdt)4+3s(d2sdt2)=0
Solution:
(dsdt)4+3s(d2sdt2)=0
(s)4+3s(sn)=0

Highest order derivative =2

Order=2

Degree= Power ofs"

Degree=1


Differential Equations Exercise Revision Exercise Question 1(ii)

Answer:
Order3,Degree1
Hint:
Check the highest order of derivate to find the order and check the power of highest derivative to find degree.
Given:
y+2y+y=0
Solution:
y+2y+y=0
Highest order derivative=3
Order=3
Degree= Power of y
Degree=1



Differential Equations Exercise Revision Exercise Question 1(iii)

Answer:
Degree 2,Order3
Hint:
Check the highest order of derivate to find the order and check the power of highest derivative to find degree.
Given:
(y)2+(y)3+(y)4+y5=0
Solution:
(y)2+(y)3+(y)4+y5=0
Highest order derivative=3
Order=3
Degree=power of(y)2
Degree=2

Differential Equations Exercise Revision Exercise Question 1(iv)

Answer:
Degree1,Order 3
Hint:
Check the highest order of derivate to find the order and check the power of highest derivative to find degree.
Given:
y+2y+y=0
Solution:
y+2y+y=0
Highest order derivative=3
Order=3
Degree=power of(y)1
Degree=1

Differential Equations Exercise Revision Exercise Question 1(v)

Answer:
Degree 1,Order2
Hint:

Check the highest order of derivate to find the order and check the power of highest derivative to find degree.
Given:
y+(y)2+2y=0
Solution:
y+(y)2+2y=0
Highest order derivative=2
Order=2
Degree=power of (y)1
Degree=1


Differential Equations Exercise Revision Exercise Question 1(vii)

Answer:
Degree is not defined, order3
Hint:
Check the highest order of derivate to find the order and check the power of highest derivative to find degree.
Given:
y+y2+ey=0
Solution:
y+y2+ey=0
Highest order derivative=3
Order=3
As the equation cannot be expressed as a polynomial of derivative
Degree is not defined



Differential Equations Exercise Revision Exercise Question 2

Answer:
Verified
Hint:
Find double derivatives of given equation and put values to verify
Given:
y=e3x
Prove that y=e3x is the solution of d2ydx2+dydx6y=0
Solution:
y=e3x
dydx=3e3xd2ydx2=(3)(3)e3xd2ydx2=9e3x
Put it in equation
d2ydx2+dydx6y=09e3x3e3x6e3x=00
Hence proved.

Differential Equations Exercise Revision Exercise Question 3(i)

Answer:
Verified
Hint:
Find the first and second derivative of given function and put values in differential equation to verify
Given:
y=ex+1
yy=0
Solution:
y=ex+1dydx=exd2ydx2=ex
Put in given differential equation,
yy=0d2ydx2dydx=0exex=00
Hence verified

Differential Equations Exercise Revision Exercise Question 3(ii)

Answer:

Verified

Hint:
Find the first and second derivative of given function and put in differential equation to verify
Given:
y=x2+2x+c
y2x2=0
Solution:
y=x2+2x+cy=dydx=2x+2y=d2ydx2=2
Put in differential equation,
y2x2=0
2x+22x2=0
0
Hence verified


Differential Equations Exercise Revision Exercise Question 3(iii)

Answer:
Verified
Hint:
Find the first and second derivative of given function and put in differential equation to verify
Given:
y=cosx+c
y+sinx=0
Solution:
y=cosx+c
y=dydx=sinx
Put in differential equation,
y+sinx=0
sinx+sinx=0
0
Hence verified

Differential Equations Exercise Revision Exercise Question 3(iv)

Answer:
Verified
Hint:
Find the first derivative of given function and put value in differential equation
Given:
y=1+x2
y=xy1+x2
Solution:
y=dydx=1211+x22x using chain rule y=x1+x2
Put in differential equation,
y=xy1+x2
LHS= x1+x2
Multiply and divide by 1+x2
=x1+x2×1+x21+x2=x1+x2(1+x2)
Where,y=1+x2
Therefore, LHS=xy1+x2 RHS..Hence verified.

Differential Equations Exercise Revision Exercise Question 3(v)

Answer:
Verified
Hint:
You must firstly solve the first derivative and put value in differential equation
Given:
y=xsinx
xy=y+xx2y2
Solution:
y=dydx=sinx+xcosx.....usingproductrule
Put in differential equation xy=y+xx2y2
LHS=xy=x(sinx+xcosx)
=xsinx+x2cosx
 RHS =y+xx2y2=xsinx+xx2(xsinx)2=xsinx+xx2(1sin2x)=xsinx+x2cos2x[sin2x+cos2x=1cos2x=1sin2x]=xsinx+x2cosx
LHS=RHS
Hence verified

Differential Equations Exercise Revision Exercise Question 3(vi)

Answer:
Verified
Hint:
Find first derivative and put value in differential equation to verify
Given:
y=a2x2
x+ydydx=0
Solution:
dydx=121a2x2(2x) using chain rule dydx=xa2x2
Put in differential equation x+ydydx
=x+a2x2(xa2x2)=xx=0
=RHS
Hence verified

Differential Equations Exercise Revision Exercise Question 4

Answer:
xdydxy=0
Hint:
You must have the knowledge about the curves and make differential equation.
Given:
y=mx
Where,m is an arbitrary constant
Solution:
y=mx
dydx=m
Substitutem=dydx iny=mx
y=dydx(x)
dydx=yx
dydxyx=0
xdydxy=0which is the required equation

Differential Equations Exercise Revision Exercise Question 5

Answer:
d2ydx2+y=0
Hint:
The number of constants is equal to the number of time we differentiate
Given:
y=asin(x+b), where a,b are constants
Solution:
y=asin(x+b)
Here, there are two constants. So, we differentiate twice
dydx=acos(x+b)
Again,
d2ydx2=asin(x+b)
d2ydx2=y
d2ydx2+y=0 which is the required differential equation.

Differential Equations Exercise Revision Exercise Question 6

Answer:
y22xyy=0
Hint:
You must know the equation of parabola
Given:
Family of parabola having vertex at origin and axis along positive direction of x-axis
Solution:
Since parabola has axis along positive x-axis , its equation is
y2=4ax … (i)
Differentiate,
2ydydx=4a
Putting value in (i)
y2=2ydydx(x)
y22xydydx=0which is the required differential equation

Differential Equations Exercise Revision Exercise Question 7

Answer:
(x29)(y)2+x2=0
Hint:
You must know about the equation of circle
Given:
Family of circles having centre on y-axis and radius 3 units
Solution:
General equation of circle is
(xa)2+(yb)2=r2
Given centre is on y-axis
Centre=(0,b) and Radius=3
Hence, our equation is
(x0)2+(yb)2=32
x2+(yb)2=9

Differentiate with respect to x ,

2x+2(yb)y=02(x+(yb)y)=0(yb)y=x(yb)=xy

Put value in (i)

x2+(yb)2=9x2+[xy]2=9x2+x2y2=9x2(y)2+x2(y)2=9x2(y)2+x2=9(y)2x2(y)29(y)2+x2=0(y)2[x29]+x2=0(x29)(y)2+x2=0


Differential Equations Exercise Revision Exercise Question 8

Answer:
xy2y=0
Hint:
You must know about the parabolas to form an equation
Given:
Family of parabola having vertex at origin and axis along the positive y-direction
Solution:
x2=4ay … (i)
Where, a is constant parameter
Differentiate with respect to x,
2x=4adydxx=2aya=x2y Put in (i) x2=4x2yyy=2yxxy=2yxy2y=0

Differential Equations Exercise Revision Exercise Question 9

Answer:
xyd2ydx2+x(dydx)2ydydx=0
Hint:
You must know about the equation of ellipses
Given:
Family of ellipses having foci on y-axis and centre at origin
Solution:
Ellipse whose foci on y-axis
x2a2+y2b2=1 [Two constants, differentiate twice]ddx[x2a2+y2b2]=ddx(1)1a2(d(x2)dx)+1b2(d(y2)dx)=01a2(2x)+1b2(2ydydx)=02xa2+2yb2(dydx)=02yb2(dydx)=2xa2yb2(dydx)=xa2yx(dydx)=b2a2yxy=b2a2

Again differentiate,



Differential Equations Exercise Revision Exercise Question 10

Answer:xyy+x(y)2yy=0
Hint: You must know about the equation of hyperbola
Given: Family of hyperbola having foci on x-axis and centre at origin
Solution: Hyperbola whose foci on y-axis
x2a2y2b2=1 [Two constants, differentiate twice]
ddx[x2a2y2b2]=ddx(1)1a2(d(x2)dx)1b2(d(y2)dx)=01a2(2x)1b2(2ydydx)=02xa22yb2(dydx)=02yb2(y)=2xa2yb2(y)=xa2yx(y)=b2a2yxy=b2a2

Again differentiate,

y[(yy)xy]x2+yxy=0.. using product and division rule of differentiation (yy+yy)xyy=0.. multiplying by x2(y2+yy)xyy=0xyy+x(y)2yy=0


Differential Equations Exercise Revision Exercise Question 11

Answer: xy11+2y1xy+x22=0
Hint: Find first and second derivative of equation put in given differential equation to verify.
Given: xy=aex+bex+x2,xd2ydx2+2dydxxy+x22x
Solution:xy=aex+bex+x2,
y+xdydx=aexbex+2x differentiating using product rule 
dydx+xd2ydx2+dydx=aex+bex+2 differentiating again w.rt to x
xy+2y2=aex+bex
xy+2y2=xyx2 [xy=aex+bex+x2xyx2=aex+bex]
xy+2yxy+x22=0
Hence proved

Differential Equations Exercise Revision Exercise Question 12

Answer: Hence verified
Hint: Find first and second derivative of given equation and put in differential equation to be verified.
Given: y=cx+2c2,2(dydx)2+xdydxy=0
Solution: y=Cx+2c2
dydx=c…differentiating w.r.t to x
d2ydx2=0..differentiating agin w.r.t to x
Now, differential equation is
2(dydx)2+xdydxy=0 L.HS=2(c)2+x(c)cx2c2=2c2+cxcx2c2=0=R.H.S
Hence verified

Differential Equations Exercise Revision Exercise Question 13

Answer: y2x2xy=a is the solution of the given differential equation
Hint: Find first and second order derivative and put values in differential equation to be verified.
Given: y2x2xy=a,(x2y)dydx+2x+y=0
Solution: y2x2xy=a
2ydydx2xxdydxy=0 differentiating wr.t to x[(x2y)dydx+2x+y]=0(x2y)dydx+2x+y=0
Thusy2x2xy=a is the solution of the given differential equation

Differential Equations Exercise Revision Exercise Question 14

Answer: Thus : y=Acosx+sinx is the solution of the given differential equation
Hint: Solve first order derivative and put values in differential equation to be verified.
Given:y=Acosx+sinx,cosxdydx+(sinx)y=1
Solution:
dydx=Asinx+cosx …Differentiating w.r.t to x,
Now, differential eq.
 L. H.S=cosx(dydx)+(sinx)ycosx(Asinx+cosx)+sinx(Acosx+sinx)Asinxcosx+cos2x+Asinxcosx+sin2xcos2x+sin2x=1=R.H.S{cos2x+sin2x=1}
Hence, Proved.

Differential Equations Exercise Revision Exercise Question 15

Answer: d3ydx87(dydx)+6y=0
Hint: The number constants is equals to the number time we differentiate.
Given: y=ae2x+be3x+cex
Solution: y=ae2x+be3x+cex
dydx=2ae2x3be3x+cexd2ydx2=4ae2x+9be3x+cexd3ydx3=8ae2x27be3x+cexd3ydx3=7(2ae2x3be3x+cex)6(ae2x+be3x+cex)d3ydx3=7(dydx)6yd3ydx37(dydx)+6y=0

Differential Equations Exercise Revision Exercise Question 16

Answer: d3ydx3=0
Hint: You must know about the equation of all parabola
Given: equation of all parabolas which have their axes parallel to y-axis is d3ydx3=0
Solution: Equation of the family of parabolas having axis parallel to y-axis is given by
(xn)2=4b(yk)..where n,k are constants
Differentiating w.r.t to x,2(xn)=4bdydx
(xn)=2bdydx
Differentiating again, 1=2b(d2ydx2)
Again Differentiating, 0=2b(d3ydx3)
i.e.d3ydx3=0

Differential Equations Exercise Revision Exercise Question 17

Answer:3y(y)2=y(1+(y)2)
Hint: The number of constant is equal to the number of times we need to differentiate.
Given: x2+y2+2ax+2by+C=0, find differential equation, not containing constants.
Solution: x2+y2+2ax+2by+C=0
Differentiate w.r.t x,
2x+2yy+2a+2by=0
Again, differentiate w.r.t x
2+2(y)2+2yy+2by=0
1+(y)2+yy+by=0..taking 2 common
b=(1+(y)2+yy)y
We have,
1+(y)2+yy+by=0
Again differentiate,
2yy+yy+yy+by=0
On substituting values of (b),
3yy+yy+((1+(y)2+yyy)y=03y(y)2+yyyy(y)2yyyy=03y(y)2=y(1+(y)2)

Differential Equations Exercise Revision Exercise Question 18

Answer: y=cos7x7cos5x5+2(x+1)5/252(x+1)8/23+C
Hint: Apply integration by parts method.
Given:dydx=sin3xcos4x+xx+1
Solution:dydx=sin3xcos4x+xx+1
dy=sin3xcos4x+xx+1)dx (integrating both sides) y=sin3xcos4xdx+xx+1)dxy=I1+I2
Now, I1=sin3xcos4xdx
=(1cos2x)(cos4x)(sinx)dx(cos2x+sin2x=1)
Let t=cosx
dt=sinxdx( diff. w.r.tx )I1=t4(1t2)dtI1=t4(t21)dt=t6t4)dtl1=t77t55+c1=cos7x7cos5x5+c1l2=xx+1)dx
Let t2=x+1
2tdt=dx( diff. w.rtx)l2=2(t21)ttdtl2=2t4t2dtI2=2t552t33+c2
y=I1+I2
y=cos7x7cos5x5+c1+2(x+1)5/252(x+1)8/23+C2y=cos7x7cos5x5+2(x+1)5/252(x+1)8/23+c(c1+C2=C)

Differential Equations Exercise Revision Exercise Question 19

Answer:y=tan1(x+2)+c
Hint: Use the formula of 1x2+1dx
Given:dydx=1x2+4x+5
Solution:dydx=1x2+4x+5dydx=1x2+4x+4+1dydx=1(x+2)2+(1)2dy=1(x+2)2+(1)2dx

Integrating both sides, we get

dy=1(x+2)2+(1)2dxy=tan1(x+21)+c1x2+a2=1atan1(xa)+cy=tan1(x+2)+c


Differential Equations Exercise Revision Exercise Question 20

Answer:x=tan1(y+1)+c
Hint: Separate y & x and than integrate both sides
Given:dydx=y2+2y+2
Solution:dydx=y2+2y+2dydx=y2+2y+1+1dydx=(y+1)2+(1)21(y+1)2+(1)2dy=dx

Integrating both sides, we get

1(y+1)2+(1)2dy=dx1x2+a2=1atan1(xa)+ctan1(y+11)+c=xx=tan1(y+1)+c


Differential Equations Exercise Revision Exercise Question 21

Answer:y+2x2=ex+c
Hint: Apply integration to find the equation
Given:dydx+4x=ex
Solution: dydx+4x=ex
dydx=ex4xdy=(ex4x)dx
integrate both sides
dy=(ex4x)dxy=ex4x22+cy=ex2x2+cy+2x2=ex+c

Differential Equations Exercise Revision Exercise Question 22

Answer: y=(x22x+2)ex+c
Hint: You must know about the formula of uvdx
Given:dydx=x2ex
Solution:dydx=x2ex
dy=x2exdxdy=x2exdx (integrate both sides) dy=x2exdx(2xexdx)dx[uvdx=uvdx(ddxuvdx)dx]y=x2ex2xexdxy=x2ex2xexdx+2exdx[uvdx=uvdx(ddxuvdx)dx]y=x2ex2xex+2ex+Cy=(x22x+2)ex+C

Differential Equations Exercise Revision Exercise Question 23

Answer:y=log|logx|+x22xsin2x4cos2x8+c
Hint: integrate both sides and then apply the formula of
Given: dydxxsin2x=1xlogx
Solution: dydxxsin2x=1xlogx
dydx=1xlogx+xsin2x
dydx=1xlogx+x2(1cos2x)[Cos2x=12sin2x]
dydx=1xlogx+x2xCos2x2
dy=(1xlogx+x2xCos2x2)dx (integrate both sides) 
dy=(1xlogxdx+12xdx12(xCos2x)dx
y=log|logx|+x22x2cos2xdx+12sin2x2dx
y=log|logx|+x22x2sin2x2+12(cos2x4)+C
y=log|logx|+x22xsin2x4cos2x8+C

Differential Equations Exercise Revision Exercise Question 24

Answer:y=log|tan2+2tanx+5|+C
Hint: Use substitution method
Given:(tan2x+2tanx+5)dydx=2(1+tanx)sec2x
Solution: (tan2x+2tanx+5)dydx=2(1+tanx)sec2x
dy=2(1+tanx)sec2x)(tan2x+2tanx+5)dx
Integrating both sides, we get,
dy=2(1+tanx)sec2x)(tan2x+2tanx+5)dx
Let,tan2x+2tanx+5=t
Differentiate w.r.t x
(2tanxsec2x+2sec2x)dx=dt2(1+tanx)sec2xdx=dtdy=1tdty=log|t|+Cy=log|tan2+2tanx+5|+C

Differential Equations Exercise Revision Exercise Question 25

Answer:cos55cos3x3+(x1)ex+C
Hint: Apply integration by parts method and formula ofuvdx
Given:dydx=sin3xcos2x+xex
Solution:dydx=sin3xcos2x+xex
dy=(sin3xcos2x+xex)dx (integrate both sides)
y=sin3xcos2xdx+xexdx
y=I1+I2
I1=sin3xcos2xdx
=(1cos2x)cos2xsinxdx
Let cosx=t
dt=sinxdx (diff. w.r.t. x)
I1=t2(1t2)dt
=t2(t21)dt=t4t2dt
I1=t55t83+c1
I1=cos5x5cos8x3+c1
Now,I2=xexdx
I2=xexdxexdx[uvdx=uvdx(ddxuvdx)dx]I2=xexex+c2=ex(x1)+c2 Now, y=I1+I2y=cos5x5cos8x3+c1+(x1)ex+c2y=cos5x5cos8x3+(x1)ex+c(c1+c2=c)

Differential Equations Exercise Revision Exercise Question 26

Answer: sinxsiny=c
Hint: You must know about trigonometric identities.
Given:tanydx+tanxdy=0
Solution:tanydx+tanxdy=0
tanxdy=tanydx1tanydy=1tanxdxcotydy=cotxdx(1tanx=cotx)
Integrating both sides
cotydy=cotxdxlog|siny|=log|sinx|+logclog|siny|+log|sinx|=logclog|sinysinx|=logc(loga+logb=logab)sinysinx=c

Differential Equations Exercise Revision Exercise Question 27

Answer: x+y+log|xy|=c
Hint: Separate x & y and then integrate
Given: (1+x)ydx+(1+y)xdy=0
Solution: (1+x)ydx+(1+y)xdy=0
(1+xx)dx=(1+yy)dy(1y+1)dy=(1x+1)dx
Integrating both sides, we get
(1y+1)dy=(1x+1)dx1ydy+dy=1xdxdxlog|y|+y=log|x|x+clog|y|+log|x|+x+y=cx+y+log|xy|=c(loga+logb=logab)

Differential Equations Exercise Revision Exercise Question 28

Answer: xtanxytan|secxsecy|+C
Hint: Use the formula of uvdx
Given: xcos2ydx=ycos2xdy
Solution:xcos2ydx=ycos2xdy
xcos2xdx=ycos2ydyysec2ydy=xsec2xdx(1cosx=secx)ysec2ydy=xsec2xdx( integrating both sides )ysec2ydytanydy=xsec2xdxtanxdx[uvdx=uvdx(ddxuvdx)dxytanylog|secy|=xtanxlog|secx|+Cxtanxytany=log|secx|log|secy|+Cxtanxytany=log|secxsecy|+C[logalogb=log(a/b)]

Differential Equations Exercise Revision Exercise Question 29

Answer: [log(secy+tany)]2=[log(secx+tanx)]2+C
Hint: Apply substitution method.
Given:cosylog|secx+tanx|dx=cosxlog|secy+tany|dy
Solution:cosylog|secx+tanx|dx=cosxlog|secy+tany|dy
log|secx+tanx|cosxdx=log|secy+tany|cosydy Integrating both sides log|secy+tany|cosydy=log|secx+tanx|Cosxdx
Let log|secy+tany|=t and log|secx+tanx|=u
(sec2y+secy tany secy+tany)dy=dt and (sec2x+secxtanxsecx+tanx)dx=du
secydy=dt and secxdx=du
$\Rightarrow\left(\frac{1}{c o s y}\right) d y=d t \text { and }\left(\frac{1}{\cos x}\right) d x=d u \ $
tdt=udu
t22=u22+c1
t2=u2+2c1
t2=u2+c(2c1=C)
[log(secy+tany)]2=[log(secx+tanx)]2+c


Differential Equations Exercise Revision Exercise Question 30

Answer: 12(logy)2+(2x2)cosx+2xsinx=C
Hint: Apply integration and then the formula of uvdx and substitution method.
Given: coscx(logy)dy+x2ydx=0
Solution:coscx(logy)dy+x2ydx=0
logyydy=x2cosecxdxlogyydy=x2sinxdx(integrating both sides)
Let log y=t
1ydy=dt( diff. w.r.t. y)⇒∵tdt=x2sinxdx12(logy)2=x2(cosx)+2x(cosx)dx[uvdx=uvdx(ddxuvdx)dx]12(logy)2=x2cosx2xcosxdx12(logy)2=x2cosx2xsinx+2sinxdx12(logy)2=x2cosx2xsinx2cosx+C12(logy)2=(x22)cosx2xsinx+C12(logy)2+(2x2)cosx+2xSinx=C

Differential Equations Exercise Revision Exercise Question 31

Answer: log|y|+log|y1|=12log|1x2|+C
Hint: you must know the rules of solving differential equation. First rearrange the values and then solve.
Given:(1x2)dy+xydx=xy2dx
Solution:(1x2)dy+xydx=xy2dx
(1x2)dy=(xy2xy)dx(1x2)dy=xy(y1)dx1y(y1)dy=x1x2dx
Integrating both sides, we get,
1y(y1)dy=x1x2dx..(I) L.H.S :1y(y1)=Ay+By11=A(y1)+By
Substituting y=1
1=A(01)+B(0)
1=B
Again substituting, y=0
1=A(01)+B(0)
1=A
A=1
Substituting values of A and B in 1y(y1)=Ay+By1, we get,
1y(y1)=1y+1y11y(y1)dy=1ydy+1y1dy=log|y|+log|y1|+C1
Now considering R.H.S of (II), we have,
x1x2dx
Here, putting 1x2=tand differentiate both sides, we get,
2xdx=dtxdx=dt2x1x2dx
121tdt12log|t|+C2[dtt=log|t|+C]12log|1x2|+C2[ where t=1x2]
Now substituting the values of 1y(y1)dy and x1x2dx in (I)
log|y|+log|y1|+C1=12log|1x2|+c2log|y|+log|y1|=12log|1x2|+C[ where C2C1=C]

Differential Equations Exercise Revision Exercise Question 32

Answer:y2logy=xsin+C
Hint: you must know the rules of solving differential equation and integrations.
Given: dydx=sinx+xcosxy(2logy+1)
Solution:dydx=sinx+xcosxy(2logy+1)
y(2logy+1)dy=(sinx+xcosx)dx
integrating both sides,
2ylogydy+1ydy=sinxdx+xcosxdx
Integrating by parts,
2logyydy2(ddy(logy)xydy)dy+ydy=cosx+xcosxdxddx(x)cosxdx2logy(y22)ydy+ydy=cosx+xsinx+cosx+cy2logy=xsinx+c

Differential Equations Exercise Revision Exercise Question 33

Answer: ey+eylog|x|+x22+C
Hint: you must know the rules of solving differential equation and integrations.
Given:x(e2y1)dy+(x21)eydx=0
Solution:x(e2y1)dy+(x21)eydx=0
x(e2y1)dy=(x21)eydxx(e2y1)dy=(1x2)eydxe2y1eydy=1x2xdx(e2yey1ey)dy=(1xx2x)dx(e2yyey)dy=(1xx)dx
Integrating both sides,
(eyey)dy=1xdxxdx(eydyeydy=1xdxxdx
ey(ey)=log|x|x22+c
ey+eylog|x|+x22+c



Differential Equations Exercise Revision Exercise Question 34

Answer:(x+c)ex+y+1=0\

Hint: you must know the rules of solving differential equation and integrations.

Given:dydx+1=ex+y

Solution:dydx+1=ex+y..........(I)

Put x + y = t and differentiate both sides. We get,

1+dydx=dtdx

Compare with equation (I),

dtdx=et

etdt=dx

Now, integrating both sides,

etdt=dx[etdt=et1]et=x+C1ex+y=x+C(x+C)ex+y+1=0


Differential Equations Exercise Revision Exercise Question 35

Answer: tan1(x+y)=x+c
Hint: you must know the rules of solving differential equation and integrations.
Given: dydx=(x+y)2
Solution:dydx=(x+y)2
Let (x+y)=u and differentiate both sides ,we get,
1+dydx=dudxdydx=dudx1(x+y)=u[(x+y)2=u2]U2=dudx1[dudx=u2]dudx=u2+1duu2+1=dx
Now, integrating both sides,
duu2+1=dxtan1=x+C[1x2+1dx=tan1x+C]tan1(x+y)=x+c where, u=x+y
Hence,tan1(x+y)=x+c

Differential Equations Exercise Revision Exercise Question 36

Answer: y=tan(x+y2)+C
Hint: you must know the rules of solving differential equation and integrations.
Given:cos(x+y)dy=dx
Solution:cos(x+y)dy=dx (dydx=1cos(x+y))
Let (x+y)=u and differentiating both sides,
1+dydx=dudxdydx=dudx11cosu=(dudx1)1=(dudx1)cosuCosududx=1+cosu
 cosu 1+cosudu=dx(1+cosu)11+cosudu=dx[1+cosu1+cosu11+cosu]du=dx[111+cosu]du=dx
[112cos2u2]du=dx[11+cosx=2cos2u2[112sec2u2]du=dx
Now, integrating both sides,
1du=1.12sec2u2du=dxUtanu2=x+c[sec2x2=2tanx2]
Put the value of u
(x+y)tan(x+y)2=x+c[u=x+y]y=tan(x+y2)+C

Differential Equations Exercise Revision Exercise Question 37

Answer:
y2x=cx2y
Hint: you must know the rules of solving differential equation and integrations.
Given: dydx+yx=y2x2
Solution: dydx+yx=y2x2
dydx=y2x2yx
dydx=y2xyx2
Put y = v x and differentiate both sides w.r.t x
dydx=v.1+xdvdxdydx=v+xdvdx
So, equation (I) becomes
v+xdvdx=v2vxdvdx=v2vvxdvdx=v22v(1v22v)dv=1xdx
Now, integrating both sides
(1(v1)21dv=1xdx[(v1)21=v2+12v1=v22v]12log(v2v)=log|x|+logc12log(v2v)=logcxlog(v2v)=2logcxlog(v2v)=logc2x2[alogx=logxa]
v2v=c2x212v=cx2[c=c]12yx=cx2 (put value of v=yx)12xy=cx2y2x=cx2y



Differential Equations Exercise Revision Exercise Question 38

Answer: 2A=log(xy)xy
Hint: you must know the rules of solving differential equation and integrations.
Given:dydx=y(xy)x(x+y)
Solution:dydx=y(xy)x(x+y)
x(x+y)dy=y(xy)dx .....................(1)
Put , y=vx and differentiate both side.
dydx=xdvdx+v
Eq. (1) becomes,
x(x+yx)(xdx+ydx)=yx(xyx)dxx2(1+v)(xdv+ydx)vx2(1v)dx=0x2[(1+v)xdv+v(1+v)dxv(1v)dx]=0x(1+v)dv+v(1+v)dxv(1v)dx=0[y(1+v)dxv(1v)dx]=x(1+v)dv[v+v2v+v2]dx=x(1+v)dv2v2dx=x(1+v)dvdxx=12v2+v2v2dvdxx=12v2+112vdv
Now, integrating both sides,
dxx=12v2dv+121vdvlog|x|+A=12v2+12+1+12log|v|log|x|+A=12v1+12logvlog|x|+A=12v+logv12 A=12v+logv12+log|x|A=12v+logv2x
Put value of v=yx

A=12(yx)+log(xy)12 A=x2y+12logxy2 A=log(xy)xy(where A is integration constant)




Differential Equations Exercise Revision Exercise Question 39

Answer:
2(x+y)4xlog|2x+2y1|=4C
Hint: you must know the rules of solving differential equation and integrations.
Given:(x+y1)dy=(x+y)dx
Solution:
(x+y1)dy=(x+y)dx
dydx=(x+y)x+y1
Putting x+y=v , we get,
1+dydx=dvdxdydx=dvdx1dvdx1=v(v1)dvdx=v(v1)+1dvdx=v+v1(v1)dvdx=2v1(v1)v12v1dv=dx
Integration both sides we get
v12v+1dv=dx122v2v1dv12v1dv=dx122v1+12v1dv12v1dv=dx12dv+1212v1dv12v1dv=dx12dv1212v1dv=dx12v14log|2v1|=x+C
12(x+y)14log|2x+2y1|=x+C2(x+y)log|2x+2y1|=4x+4C2(x+y)4xlog|2x+2y1|=4C

Differential Equations Exercise Revision Exercise Question 40

Answer: ycosecx+cotx=c
Hint: you must know the rules of solving differential equation and integrations.
Given: dydxycotx=cosecx
Solution:dydxycotx=cosecx
The above equation is in form of dydx+py=q

Where p = -cot x and q = cosec x

Integrating factor = epx

Considering pdx

pdx=cotxdx=log|sinx|[elogx=x]epdx=elog|sinx|=elog(sinx)1=sinx1=1sinx=cosecx

 Integrating factor I.F =cosecx

Now, General solution is,

y(I.F)=q(I.F)dx+Cycosecx=cosecxcosecxdx+cycosecx=cosecx2dx+cycosecx=cotx+Cycosecx+cotx=c




Differential Equations Exercise Revision Exercise Question 41

Answer:ycosx=cos2x2+c
Hint: you must know the rules of solving differential equation and integrations.
Given:dydxytanx=2sinx
Solution:dydxytanx=2sinx
dydx+(tanx)y=2sinx
The above equation look like,
dydx+py=q
Where p = -tan x and q = -2 sin x
Integrating factor = epx
=etanxdx
=etanxdx
we have tanxdx=log(secx)+c I. F=elog(secx) I.F =elog(1secx) I.F =1secx[elogx=x] I. F=Cosx

Hence, now the solution of differential equation is,

y(I.F)=(q×IF)dx+Cy(Cosx)=(2sinxcosx)dx+CyCosx=sin2xdx+Cycosx=cos2x2+c


Differential Equations Exercise Revision Exercise Question 42

Answer: ycosx=ex+C
Hint: you must know the rules of solving differential equation and integrations.
Given:dydxytanx=exsecx
Solution:dydxytanx=exsecx
Comparing with,
dy dx +py=q (I) we get, P=tanx and q=exsecx Now, I.F =etanxdx=elog|(secx)|=elog(1secx)=elog(cosx)[elogx=x]=Cosx
So, the solution is,
yI.F=(qxIF)dx+CyCosx=(cosxexsecx)dx+C=cosxex1cosxdx+C=exdx+c[exdx=ex+C]yCosx=ex+C

Differential Equations Exercise Revision Exercise Question 43

Answer: ycosx=ex2(sinx+cosx)+C
Hint: you must know the rules of solving differential equation and integrations.
Given:dydxytanx=ex
Solution:dydxytanx=ex
Compare with dydx+py=q, we get,
P=tanx and q=ex Now, I.F=etanxdx=elog|(secx)|=elog(1sec)[elogx=x] I. F=1secx=Cosx
Now the solution is,
yIF=(q×I.F)dx+CyCosx=(cosxex)+C=Cosexdxex(sinx)dx+C=excosx+sinxexcosxexdx+C=excosx+exsinx2+CyCosx=ex2(sinx+cosx)+C

Differential Equations Exercise Revision Exercise Question 44

Answer: xy=tan1x+C
Hint: you must know the rules of solving differential equation and integrations.
Given:(1+y+x2y)dx+(x+x3)dy=0
Solution:(1+y+x2y)dx+(x+x3)dy=0
x(1+x2)dy=[(1+y(1+x2)]dxdydx=1y(1+x2)x(1+x2)dydx=1xy1x(1+x2)dydx+1xy=1x(1+x2)
Comparing with dydx+py=q , we get,
P=1x and q=1x(1+x2) I.F =e1xdx=elogx=x[elogx=x]
The solution is,
yIF=1x(1+x2)xdxyx=1(1+x2)dxyx=tan1x+C

Differential Equation Exercise Revision Exercise (RE) Question 45

Answer : y=ce2tan1x+12
Hint: you must know the rules of solving differential equation and integrations.
Given: (x2+1)dy+(2y1)dx=0
Solution : (x2+1)dy+(2y1)dx=0
(x2+1)dy=(2y1)dx
dydx=(12y)(1+x2)dydx=1(1+x2)2y(1+x2)dydx+2y(1+x2)=1(1+x2)
Comparing with dydx+py=q, we get,
P=2(1+x2) and q=11+x2 Now, I.F =epdx
=e2dx1+x2=e2tan1x
The solution is,
yxI.F=Q×IF+Cye2tan1x=e2tan1x1+x2dx
Put tan1x=t, differentiating both side,
11+x2dx=dtyne2tan1x=e2tdtye2tan1x=12e2t+cm[eatdx=1aeax+c]
 y e2tan1x=12e2tan1x+cy=ce2tan1x+12

Differential Equation Exercise Revision Exercise (RE) Question 46

Answer : y=log(cosec2xcot2x)y7+c
Hint: you must know the rules of solving differential equation and integrations.
Given: ysec2x+(y+7)tanxdydx=0
Solution : ysec2x+(y+7)tanxdydx=0
(y+7)tanxdydx=ysec2x(y+7)dy=ysec2xtanxdx(y+7)ydy=1cos2xxcosxsinxdx
(1+7y)dy=1cosxsinxdx(1+7y)dy=22cosxsinxdx(1+7y)dy=2sin2xdx
Now, Integrating both sides,
1dy+7ydy=2sin2xdxy+7log|y|=2cosec2xdxy+7log|y|=2[12log|cosec(2x)cot(2x)|]+c
y+logy7=log|cosec(2x)cot(2x)|+cy=log|cosec(2x)cot(2x)|logy7+cy=log(cosec2xcot2x)y7+c

Differential Equation Exercise Revision Exercise (RE) Question 47

Answer : y+c=a2[log|x|log|x+2a|]
Hint: you must know the rules of solving differential equation and integrations.
Given: (2ax+x2)dydx=a2
Solution : (2ax+x2)dydx=a2
dy=(a2(2ax+x2))dx
split
dy=(a2x(x+2a))dxdy=a2(1x1x+2a)dx
Integrating both sides,
dy=a2[1xdx1x+2adx]y+c=a2[log|x|log|x+2a|]


Differential Equation Exercise Revision Exercise (RE) Question 48

Answer : c=4x3y2y4
Hint: you must know the rules of solving differential equation and integrations.
Given: (x32y3)dx+3x2ydy=0
Solution : (x32y3)dx+3x2ydy=0
3x2ydy+(x32y3)dx=03x2ydy=(2y3x3)dx
dydx=2y3x33x2ydxdy=3x2yx32y3=3xy+2y2x2=32(yx)2xy
Put yx=y, differentiate both sides,
y=vx
dydx=v+xdvdxv+xdvdx=32v21vxdvdx=3v2v4+v2v31
Integrating both sides,
dxx=2v313v2v4+vdVdxx=2v314v2v4 d V
put, 4v2v4=t
(48v3)dv=dt4(2v31)dv=dtdxx=14dttlogx+logc=14logt
log(cx)=logt14t14=cx
Put value of t
(4v2v4)14=cx
put value of v=yx
(4yx2y4x4)14=cx(4x3y2y4x4)14=cxx(4x3y2y4)14=cx[(x4)14=x1]
1(4x3y2y4)14=c1c=(4x3y2y4)14c4=4x3y2y4c=4x3y2y4

Differential Equation Exercise Revision Exercise (RE) Question 49

Answer : c=xetan1yx
Hint: you must know the rules of solving differential equation and integrations.
Given: x2dy+(x2xy+y2)dx=0
Solution : x2dy+(x2xy+y2)dx=0
x2dy=(xyx2y2)dxdydx=xyx2y2x2
putting y=vx and differentiate,
dydx=v+xdvdxv+xdvdx=xyx2y2x2 [put value of dydx]v+xdvdx=v1v2
xdvdx=v1v2vxdvdx=1v2dv1+v2=1xdx
Integrating both sides,
dv1+v2dv=1xdxtan1v=log|x|+logc[11+x2dx=tan1x]tan1v=logcx
etan1v=cx[logx=a,x=ea]c=xetan1yx [put value of v=xy]

Differential Equation Exercise Revision Exercise (RE) Question 50

Answer : x=c1(yb)(bx+1)
Hint: you must know the rules of solving differential equation and integrations.
Given: (yx)dydx=b(1+x2dydx)
Solution : (yx)dydx=b(1+x2dydx)
yb=(bx2+x)dydx(1yb)dy=(1bx2+x)dx
Integrating both sides,
1ybdy=1bx2+xdx1ybdy=1b1x2+xbdx1ybdy=1b(1x2+xb+14b214b2)dx
(1yb)dy=1b(1(x+12b)2(12b)2)dx
log|yb|=12x12 b×blog|x+12b12bx+12b+12b|+logc[1xdx=log|x| and 1(x+a)2a2=log|x+aax+a+a|]log|yb|=log|bxbx+1|+logc
yb=cbxbx+1cbx=(yb)(bx+1)x=c1(yb)(bx+1)[ where 1cb=c1]

Differential Equation Exercise Revision Exercise (RE) Question 51

Answer : y=113(2sin3x3cos3x)+Ce2x
Hint : You must know the rules of solving differential equation and integration.
Given : dydx+2y=sin3x
Solution : dydx+2y=sin3x
Compare with,
dydx+Py=Q
where, P=2 and Q=sin3x
Therefore,
 Integrating factor = I.F =epdx=e2dx=e2x
The solution is,
y×I.F=(Q×I.F)dx+Cye2x=e2x×sin3xdx+Cye2x=I+C ......(i)
Where, I=e2xsin3xdx ......(ii)
Apply integrating by parts,
I=e2xsin3xdx[de2xdxsin3xdx]dxI=e2xcos3x3+23e2xcos3xdxI=e2xcos3x3+23[e2xcos3xdx{de2xdxcos3xdx}dx]
I=e2xcos3x3+23[e2xsin3x323e2xsin3xdx]I=[29e2xsin3xe2xcos3x3]49I
I+49I=e2x(29sin3xcos3x3)139I=e2x(29sin3xcos3x3)I=913e2x(29sin3xcos3x3)I=113e2x(2sin3x3cos3x)+C .....(iii)
From (i) and (iii), we get,
ye2x=e2x13(2sin3x3cos3x)+Cy=113(2sin3x3cos3x)+Ce2x Is required solution.

Differential Equation Exercise Revision Exercise (RE) Question 52

Answer : y=4(x1)+cex
Hints : You must know the rules of solving differential equation and integration.
Given : dydx+y=4x
Solution : dydx+y=4x .....(i)
Compare with, dydx+py=Q
Where, P=1,Q=4x
Therefore,
 I.F =ePdx=edx=ex
Hence, the solution is ,
y×I.F=(I.F×Q)dx+cyex=ex4xdx+c
Integrating by parts,
yex=4xexdx4[ddx(x)exdx]+cyex=4xex4exdx+cyex=4xex4ex+c
yex=4(x1)ex+cy=4(x1)+cex is required solution

Differential Equation Exercise Revision Exercise (RE) Question 53

Answer : 141(4sin4x+5cos4x)+ce5x
Hint : You must know the rules of solving differential equation and integration.
Given :dydx+5y=cos4x
Solution :dydx+5y=cos4x
Compare with,
dydx+Py=Q
When P=5,Q=cos4x
Therefore,
I.F=ePdx=e5dx=e5x
The solution is ,
y×I.F=(I.F×Q)dx+cye5x=e5x×cos4xdx+cyex=I+c ....(i)
Where,
I=e5xcos4xdxI=e5xcos4xdx[de5xdxcos4xdx]dxI=e5xsin4x454e5xsin4xdx
I=e5xsin4x454[e5xsin4xdx{de5xdxsin4xdx}dx]I=e5xsin4x454[e5xcos4x4+54e5xcos4xdx]I=e5xsin4x4+516e5xcos4x2516e5xcos4xdx
I=e5x16(4sin4x+5cos4x)2516I2516I+I=e5x16(4sin4x+5cos4x)4116I=e5x16(4sin4x+5cos4x)I=e5x41(4sin4x+5cos4x)
Therefore, required solution is
ye5x=e5x41(4sin4x+5cos4x)+cy=141(4sin4x+5cos4x)+ce5x

Differential Equation Exercise Revision Exercise (RE) Question 54

Answer : tanyx=log|x|+c
Hint : You must know the rules of solving differential equation and integration
Given : xdydx+xcos2(yx)=y
Solution : xdydx+xcos2(yx)=y
dydx+cos2(yx)=yxdydx=yxcos2(yx)
Putting y=vx and differentiate
Therefore, dydx=v+xdvdx
Therefore,
v+xdvdx=vcos2(v)1cos2vdv=dxxsec2vdv=1xdx
Integrating both sides,
sec2dv=1xdxtanv=log|x|+ctanyx=log|x|+c

Differential Equation Exercise Revision Exercise (RE) Question 55

Answer : yetanx=etanx(tanx1)+c
Hint : You must know the rules of solving differential equation and integration
Given : cos2xdydx+y=tanx
Solution : cos2xdydx+y=tanx
dydx+ycos2x=tanxcos2xdydx+(sec2x)y=(tanx)sec2x
Comparing with ,
dydx+Py=Q
Where, P=sec2x,Q=(tanx)(sec2x)
Now,
 I.F =esec2xdx=etanx
So, the solution is
y×I.F=(I.F×Q)dx+cy×etanx=etanx×(tanx)(sec2x)dx+cyetanx=I+C
Now,
I=tanx(sec2x)×etanxdx
Integrating by parts,
 Put t=tanxdt=sec2xdx
Therefore,
I=(t×et)dtI=t×etdt(dtdt×etdt)dt=tetetdt=tetet
Therefore,
I=tanxetanxetanx=etanx(tanx1)
Hence, the required solution is
yetanx=etanx(tanx1)+c




Differential Equation Exercise Revision Exercise (RE) Question 56

Answer : xysecx=tanx+c
Hint : You must know the rules of solving differential equation and integration
Given : xcosxdydx+y(xsinx+cosx)=1
Solution :
xcosxdydx+y(xsinx+cosx)=1dydx+y(xsinxxcosx+cosxxcosx)=1xcosxdydx+(tanx+1x)y=1xcosx
Comparing with, dydx+Py=Q
Where, P=tanx+1x,Q=1xcosx
Therefore,
I.F=e(tanx+1x)dx=elog|secx|+log|x|=elog|xsecx|=|xsecx| [elogx=x]
So the solution is,
y×I.F=(I.F)×(Q)dx+Cxysecx=xsecx×1xcosxdx+c=sec2x+cxysecx=tanx+c [sec2xdx=tanx]

Differential Equation Exercise Revision Exercise (RE) Question 57

Answer : xetan1y=tan1y+c
Hint : You must know the rules of solving differential equation and integration
Given : (1+y2)+(xetan1y)dydx=0
Solution : (1+y2)+(xetan1y)dydx=0
dxdy=etan1yx1+y2dxdy+x1+y2=etan1y1+y2
Comparing with,
dxdy+Px=Q,weget
P=11+y2,Q=etan1y1+y2
Now,  I.F =e11+y2dy=etan1y
So, the solution is
x×I.F=I.F×Qdy+cx×etan1y=etan1y1+y2×etan1ydy+cx×etan1y=11+y2dy+cxetan1y=tan1y+c

Differential Equation Exercise Revision Exercise (RE) Question 58

Answer : x=(1+1y)+ce1y
Hint : you must know the rules of solving differential equation and integration
Given : y2+(x+1y)dydx=0
Solution : y2+(x+1y)dydx=0
dydx=y3xy+1dxdy=xy+1y3dxdy=xy21y3dxdy+xy2=1y3
Comparing with, dxdy+Px=Q
Where , P=1y2,Q=1y3
Now,
 I. F=e1y2dy=e1y
therefore the solution is
x×I.F= I.F ×Qdy+cxe1y=e1y1y3dy+cxe1y=I+c
Putting t=1y, and differentiating both sides
dt=1y2dy
Applying integration both side,
I=(t×et)dtI=t×etdt(dtdt×etdt)dt=tet+etdt=tetet
Therefore,
I=1ye1ye1y[t=1y]I=e1y(1+1y)
Hence, required solution is ,
xe1y=e1y(1+1y)+cx=(1+1y)+ce1y

Differential Equation Exercise Revision Exercise (RE) Question 59

Answer : y=cosx2cos2x
Hint : you must know the rules of solving differential equation and integration
Given : 2cosxdydx+4ysinx=sin2x, y=0,x=Π3
Solution : 2cosxdydx+4ysinx=sin2x
dydx+4ysinx2cosx=2sinxcosx2cosxdydx+2ytanx=sinx
Comparing with, dydx+Py=Q , we get
Where, P=2tanx,Q=sinx
Now,
 I.F =e2tanxdx=e2log|secx|=sec2x
So, the solution is,
y× I.F =(Q×I.F)dx+cysec2x=sinxsec2xdx+cysec2x=tanxsecxdx+c
ysec2x=secx+cy=cosx+Ccos2x now x=π3,y=0
Therefore,
0=cosπ3+Ccos2π30=12+C(14)
C=2
Putiing value of C,
y=cosx2cos2x

Differential Equation Exercise Revision Exercise (RE) Question 60

Answer : xetan1y=etan1y(tan1y1)+c
Hint : You must know the rules of solving differential equation and integration
Given : (1+y2)dx=(tan1yx)dy
Solution : (1+y2)dx=(tan1yx)dy
dxdy=tan1yx1+y2dxdy+x1+y2=tan1y1+y2
Comparing with, dxdy+Px=Q,
P=11+y2,Q=tan1y1+y2 Now  I. F=e11+y2dy=etan1y
So, the solution is ,
x×I.F=(I.F×Q)dy+cxetan1y=tan1y1+y2×etan1ydy+cxetan1y=I+C
Now, I=tan1y1+y2×etan1ydy
 Put μ tan 1y=t and differentiate, 11+y2dy=dtI=ettdt
Integrating by parts,
I=t×etdt{dtdtetdt}dtI=tetetdtI=tetet
I=tan1yetan1yetan1yI=etan1y(tan1y1)
By putting value of I
We get the required solution,
xetan1y=etan1y(tan1y1)+c

Differential Equation Exercise Revision Exercise (RE) Question 61

Answer : ysecx=xn+1n+1+c
Hint : You must know the rules of solving differential equation and integration
Given : dydx+ytanx=xncosx
Solution : dydx+ytanx=xncosx
Comparing with, dydx+Py=Q
P=tanx,Q=xncosx
Now,
 I.F =etanxdx=elog(secx)=secx
So the solution is ,
y× I.F =(Q×I.F)dx+cysecx=xncosxsecxdx+cysecx=xndx+cysecx=xn+1n+1+c

Differential Equation Exercise Revision Exercise (RE) Question 62

Answer : x2+y2+2x4y+c=0
Hint : You must know the rules of solving differential equation and integration
Given : dydx=x+12y,y2
Solution : dydx=x+12y
(2y)dy=(x+1)dx
Integrating both sides,
(2y)dy=(x+1)dx2yy22=x22+x+c1x22+x+c12y+y22=0
x2+2x+y24y+2c1=0x2+y2+2x4y+c=0[2c1=c]

Differential Equation Exercise Revision Exercise (RE) Question 63

Answer : y=12x2+1
Hint : You must know the rules of solving differential equation and integration
Given : dydx=4xy2y=1,x=0
Solution : dydx=4xy2
1y2dy=4xdx
Integrating both sides ,
1y2dy=4xdx1y=2x2+c
Now, x=0,y=1
Therefore, 1=0+c
c=1
Put value of c,
1y=2x211y=2x2+1y=12x2+1

Differential Equation Exercise Revision Exercise (RE) Question 64 (i)

Answer : y=2tanx2x+c
Hint : You must know the rules of solving differential equation and integration
Given : dydx=1cosx1+cosx
Solution : dydx=1cosx1+cosx
dydx=2sin2x22cos2x2dydx=2tan2x2
dy=(tan2x2)dxdy=(sec2x21)dx
Integrating both sides,
dy=(sec2x21)dxdy=sec2x21dxy=2tanx2x+c

Differential Equation Exercise Revision Exercise (RE) Question 64 (ii)

Answer : y=2sin(x+c)
Hint : You must know the rules of solving differential equation and integration
Given : dydx=4y2,2<y<2
Solution : dydx=4y2,2<y<2
14y2dy=dx
Integrating both sides,
14y2dy=dxsin1y2=x+cy2=sin(x+c)y=2sin(x+c)

Differential Equation Exercise Revision Exercise (RE) Question 64 (iii)

Answer : tan1y=x+x33+c
Hint : You must know the rules of solving differential equation and integration
Given : dydx=(1+x2)(1+y2)
Solution : dydx=(1+x2)(1+y2)
11+y2dy=(1+x2)dx
Integrating both sides,
11+y2dy=(1+x2)dx[11+y2dy=tan1y+c]
tan1y=x+x33+c



Differential Equation Exercise Revision Exercise (RE) Question 64 (iv)

Answer : y=ecx
Hint : You must know the rules of solving differential equation and integration
Given : ylogydxxdy=0
Solution : ylogydxxdy=0
ylogydx=xdy
1xdx=1ylogydy1ylogydy=1xdx
Integrating both sides,
1ylogydy=dxx
Put logy=t and differentiating,
1ydy=dt
Therefore,
1tdt=1xdxlog|t|=logx+logclog(logy)=logx+logc
log(logy)=logcxlogy=cxy=ecx

Differential Equation Exercise Revision Exercise (RE) Question 64 (v)

Answer : y=xsin1x+1x2+c
Hint : You must know the rules of solving differential equation and integration
Given : dydx=sin1x
Solution : dydx=sin1x
dydx=sin1xdx
Integrating both sides
dy=1×sin1xdx
Integrating by parts,
Put dy=sin1x1dx[ddx(sin1x)1dx]dxy=xsin1xx1x2dxt2=1x2, and differentiate we get 
2tdt=2xdxtdt=xdx
Therefore,
y=xsin1x+dty=xsin1x+t+cy=xsin1x+1x2+c

Differential Equation Exercise Revision Exercise (RE) Question 64 (vi)

Answer : y=1+Cex
Hint : You must know the rules of solving differential equation and integration
Given : dydx+y=1
Solution : dydx+y=1
dydx=1y11ydy=dx
Integrating both sides
1(1y)dy=dx1y1dy=dx1y1dy=dx
log|y1|=x+logclog|y1|logc=xlog|y1c|=x
y1c=exy=1+Cex

Differential Equation Exercise Revision Exercise (RE) Question 65 (i)

Answer : y=12log(x21x2)12log(34)
Hint : You must know the rules of solving differential equation and integration
Given : x(x21)dydx=1,y=0 where x=2
Solution : x(x21)dydx=1
dydx=1x(x21)dy={1x(x21)}dx
Integrating both sides,
dy={1x(x21)}dxy=1x(x+1)(x1)dx+c[a2b2=(a+b)(ab)]
Let,
1x(x+1)(x1)=Ax+Bx+1+Cx11=A(x+1)(x1)+Bx(x1)+Cx(x+1)1=A(x21)+B(x2x)+C(x2+x)1=x2(A+B+C)+x(B+C)A
Comparing both sides,
A=1B+C=0A+B+C=0
Therefore, A=1,B=12,C=12
Therefore,
1x(x+1)(x1)=1x+12(x+1)+12(x1)
Now,
y=(1x+12(x+1)+12(x1))dx+cy=log|x|+12log|x+1|+12log|x1|+cy=12log|x+1|+12log|x1|log|x|+c
Given, y(2)=0
Therefore, y=0,x=2
0=12log|2+1|+12log|21|log|2|+cc=log|2|12log(3)[log1=0]
Therefore the solution is ,
y=12log|x1|+12log|x+1|log|x|+log(2)12log(3)2y=log|x1|+log|x+1|2log|x|+2log(2)log32y=log|x1|+log|x+1|logx2+log(22)log3
2y=log(x1)(x+1)x2[log(3)log(4)]y=12logx21x212log(34)


Differential Equation Exercise Revision Exercise (RE) Question 65 (ii)

Answer : cos(y1x)=a
Hint : You must know the rules of solving differential equation and integration
Given : cos(dydx)=a
Solution : cos(dydx)=a
dydx=cos1ady=cos1adx
Integrating both sides,
dy=cos1adxdy=cos1adxy=xcos1a+c
Now, x=0,y=1
Therefore, 1=0+C
C=1
Hence,
y=xcos1a+1cos(y1x)=a

Differential Equation Exercise Revision Exercise (RE) Question 65 (iii)

Answer : y=secx
Hint : You must know the rules of solving differential equation and integration
Given : dydx=y tan x,y=1 where ,x=0
Solution : dydx=y tan x
1ydy=tanxdx
Integrating both sides,
1ydy=tanxdxlogy=log|secx|+C
Now, x=0,y=1
Therefore,
log1=log1+CC=0
Put value of c ,
logy=log|secx|+0logy=log|secx|y=secx

Differential Equation Exercise Revision Exercise (RE) Question 66 (i)

Answer : log|x2+xy+y2|=23tan1(x+2y3x)+c
Hint : You must know the rules of solving differential equation and integration
Given : (xy)dydx=x+2y
Solution :
STEP : 1
dydx=x+2yxy
STEP : 2
 Put f(x)=dydx, find f(λx,λy)dydx=x+2yxy Put f(x,y)=x+2yxy Find f(λx,λy)
f(λx,λy)=λx+2(λy)λxλy=λ(x+2y)λ(xy)=x+2yxy .....(i)
=f(x,y)
STEP : 3
Puty=vx
dydx=d(vx)dx So dydx=dvdx×x+vdxdxdydx=xdvdx+v Put dydx and yx in (i)
dvdxx+v=x+2vxxvxdvdxx+v=x(1+2v)x(1v)dvdxx=(1+2vv(1v))1v
dvdxx=1+2vv+v21vdvdxx=v2+v+11vdvdxx=(v2+v+1v1)
dv(v1v2+v+1)=dxx(v1)v2+v+1dv=dxx(v1)v2+v+1dv=log|x|+C
 Use v2+v+1=v2+12×2v+(12)(22)+1(12)22(v+12)2+114(v+12)2+34
 Put v2+v+1=(v+12)2+34v1=v+12121=(v+12)32
(v+12)32(v+12)2+34dv=log|x|+Cv+12(v+14)2+34dv321(v+12)2+34dv=log|x|+C
So, our equation become
I132I2=log|x|+c ....(ii)
Solving I1,
I1=v+12(v+12)2+34dv Put (v+12)2+34=t
Diff.w.r.t.v
d((v+12)2+34)dv=dtdv2(v+12)=dtdvdv=dt2(v+12)
Put value of v and dv in I1
I1=v+12t×dt2(v+12)=12dtt=12log|t|=12log|(v+12)2+32|12log|v2+v+1|
Solving I1
I2=dv(v+12)2+34=dv(v+12)2+(32)2 Using dxx2+a2=1atan1xa+c
 Where x=v+12 and a=32=132tan1(v+12)32=23tan1(2v+13)
From (ii)
I132I2=log|x|+C12log|v2+v+1|32×23tan1(2v+13)=log|x|+C12log|v2+v+1|3tan12v+13=log|x|+C
Replace v by yx
12log|(yx)2+yx+1|3tan1(2yx+13)=log|x|+C12log|y2x2+yx+1|+log|x|=3tan1(2y+x3x)+C
Multiply both side by 2
log|y2x2+yx+1|+log|x|2=23tan1(2y+x3x)+2C Put 2c=clog(x2y2x2+x2yx+x2)=23tan1(x+2y3x)+Clog|y2+xy+x2|=23tan1(x+2y3x)+C



Differential Equation Exercise Revision Exercise (RE) Question 66 (ii)

Answer : sinyx=log|x|+c
Hint : You must know the rules of solving differential equation and integration
Given : xcos(yx)dydx=ycos(yx)+x
Solution : given differential equation xcos(yx)dydx=ycos(yx)+x
dydx=ycosyx+xxcosyx.(i)
It is a homogeneous differential equation
Put y=vxdydx=v+xdvdx
Eqn (i) becomes
v+xdvdx=vxcosv+xxcosvxdvdx=vcosv+1cosvvxdvdx=vcosv+1vcosvcosv
xdvdx=1cosvcosvdv=dxx
Integrating both sides
sinv=log|x|+csinyx=log|x|+c

Differential Equation Exercise Revision Exercise (RE) Question 66 (iii)

Answer : cy=log|yx|1
Hint : You must know the rules of solving differential equation and integration
Given : ydx+xlog(yx)dy2xdy=0
Solution : ydx+xlog(yx)dy2xdy=0
dy(xlog(yx)2x)=ydxdydx=yxlog(yx)2xdydx=yx(2log(yx))dydx=yx2log(yx)
Step : 2
Putting f(x,y)=dydx and finding (f(λx,λx))
f(x,y)=yx2log(yx)f(λx,λy)=λyλx2log(λyλx)=yx2log(yx)f(λx,λy)=f(x,y)
Step : 3
Solve by y=vx
Put y=vx
Diff w.r.t.x
dydx=xdvdx+vdxdxdydx=xdvdx+v
Put value of dydx and y=vx in eqn(i) 
dydx=yx2logyxxdvdx=v2logvvxdvdx=v2v+vlogv2logv
xdvdx=vlogvv2logv2logvvlogvvdv=dxx
Integrate both sides
2logvvlogvvdv=dxx1+1logvv(logv1)dv=logx+logc1v(logv1)dv1vdv=logx+logcdvv(logv1)logv=logx+logc
Put t=logv1
dt=1vdv
So our equation
dttlogv=logx+logc
Put value of t
log(logv1)logv=logx+logclog(logv1)=logx+logc+logvlog(logv1)=log(xcv) put v=yx
log(logyx1)=logxcyxlog(logyx1)=logcylogyx1=cycy=logyx1



Differential Equation Exercise Revision Exercise (RE) Question 66 (iv)

Answer : 12(sinxcosx)+cex
Hint : You must know the rules of solving differential equation and integration
Given : dydxy=cosx
Solution : differential equation is in form of
dydx+Py=QP=1,Q=cosx I.F =ePdx=e1dx=ex
Solution is
y×IF=Q×IFdx+cyex=excosx+c...(i) let I=excosxdx
Integrate by f(x)g(x)dx=f(x)g(x)dx{f(x)g(x)dx}dx
 Take f(x)=cosx and g(x)=exI=cosxexdx{(sinx)exdx}dxI=excosxsinx(ex)dx
I=excosxexsinxdxI=excosx{sinxexdx(cosxexdx)dx}
I=excosx(exsinx)+excosxdxI=excosx+exsinxexcosxdxI=ex(sinxcosx)I2I=ex(sinxcosx)I=ex2(sinxcosx)
From (i)
yex=excosx+cyex=ex2(sinxcosx)+cy=12(sinxcosx)+cex

Differential Equation Exercise Revision Exercise (RE) Question 66 (v)

Answer : y=x24+cx2
Hint : You must know the rules of solving differential equation and integration
Given : xdydx+2y=x2,x0
Solution : xdydx+2y=x2
Divide both sides by x
xxdydx+2yx=x2xdydx+2yx=x
Differentiate equation in the form ,
dydx+Py=Q
Where, P=2x,Q=x
 I.F =e2xdx
 I. F=e2logx(logx=logxn) I. F=elogx2(elogx=x)
 I. F=x2
Solution of differential equation is
y×I.F=Q×I.Fdx+Cyx2=x×x2dx+Cyx2=x3dx+C
x2y=x44+Cy=x24+Cx2

Differential Equation Exercise Revision Exercise (RE) Question 66 (vi)

Answer : ye2x=e2x5(2sinxcosx)+c
Hint : You must know the rules of solving differential equation and integration
Given : dydx+2y=sinx
Solution : put in form dydx+Py=Q
dydx+2y=sinx ...(i)
Step : find P and Q
Compare (i) with dydx+Py=Q
P=2Q=sinx
Find integration factor I.F
 I.F =ePdx I.F =e2dx=e2x So I.F=e2x
Step : 4
y×I.F=Q×I.Fdx+C
Putting value
ye2x=sinxe2xdx+c Let I=sinxe2xdx
Solving I
I=sinxe2xdx=sinxe2xdx[ddxsinxe2xdx]dx=sinxe2x2cosxe2x2dx
=12sinxe2x12[cosxe2xdx{sinxe2x2dx}dx]=12sinxe2x12(cosxe2x2)14sinxe2xdxI=12sinxe2x14cosxe2x14I
I+14I=14(2sinxe2xcosxe2x)5I4=e2x4(2sinxcosx)I=e2x5(2sinxcosx)
Substituting I in eq (ii)
ye2x=e2x5(2sinxcosx)+c

Differential Equation Exercise Revision Exercise (RE) Question 66 (vii)

Answer : y=e2x+ce3x
Hint : You must know the rules of solving differential equation and integration
Given : dydx+3y=e2x
Solution : put in form dydx+Py=Q
dydx+3y=e2x
Step : 2
Find P and Q by comparing , P=3,Q=e2x
Step : 3
Find integrating factor
 I.F =ePdx I.F =e3dxI.F=e3x
Step : 4
Solution of equation
y×I.F=Q×I.Fdx+c
Putting values
y×e3x=e2x+3xdx+cye3x=exdx+cye3x=exdx+c
divide by e3x
y=e2x+ce3x

Differential Equation Exercise Revision Exercise (RE) Question 66 (viii)

Answer : xy=x44+c
Hint : : integrate by applying integration of yn
Given : dydx+yx=x2
Solution : dydx+yx=x2
Differential equation is in the form of
dydx+Py=Q
P=1x , Q=x2
Putting I.F
 I.F =ePdx I.F =e1xdx I.F =elogx I.F =x
Solution is
y×I.F=(Q×I.F)dx+cyx=x2×xdx+cyx=x3dx+cyx=x44+c

Differential Equation Exercise Revision Exercise (RE) Question 66 (ix)

Answer : y(secx+tanx)=secx+tanxx+c
Hint : integrate by applying integration of sec x and tan x
Given : dydx+secx(y)=tanx
Solution : differential equation is of the form
dydx+Py=Q Where LP=secx,Q=tanx
Finding integrating factor
I.F=ePdxI.F=esecxdxIF=elog|secx+tanx| I. F=secx+tanx
Solution is
y×I.F=(Q×I.F)dx+cy(secx+tanx)=tanx(secx+tanx)dx+cy(secx+tanx)=tanxsecxdx+tan2dx+c
y(secx+tanx)=tanxsecxdx+(sec2x1)dx+cy(secx+tanx)=secx+tanxx+c

Differential Equation Exercise Revision Exercise (RE) Question 66 (x)

Answer : y=x216(4log|x|1)+cx2
Hint : integrate by applying integration of xn and logx
Given : xdydx+2y=x2logx
Solution : convert the given differential equation is of the form dydx+Py=Q
xdydx+2y=x2logx
Divide both sides by x
dydx+2yx=xlogx Now P=2x,Q=xlogx
Differentiate
 I. F=ePdx I. F=e2xdx I.F =e21xdx I. F=e2logx I. F=elogx2=x2
Solution is
y× I.F =(Q×I.F)dx+cyx2=xlogx×x2dx+cyx2=logx×x3+c
 Use f(x)×g(x)dx=f(x)g(x)dx{f(x)g(x)dx}dxyx2=logxx3dx{ddxlogxx3dx}dxyx2=logx(x44)1x(x44)dx+c
yx2=logx×x44x34dx+cyx2=x4logx4x416+c
y=x4logx4x2x416x2+cx2y=x2log|x|4x216+cx2y=x216(4log|x|1)+cx2

Differential Equation Exercise Revision Exercise (RE) Question 66 (xi)

Answer : ylogx=2x(1+logx)+c
Hint : you integrate by integrating xn
Given : xlogxdydx+y=2xlogx
Solution :  put in form dydx+Py=Q
xlogxdydx+y=2xlogx
Divide bot side by (xlogx)
dydx+yxlogx=2xlogx×1xlogxdydx+(1xlogx)y=2x2 ...(i)
By comparing (i) with dydx+Py=Q
P=1xlogx and Q=2x2 I. F=ePdx I. F=e1xlogxdx Let t=logx
dt=1xdxI.F=e1tdtIF=elog|t| I.F =|t|=log|x|
y× I.F =Q× I.F dx+cy×logx=2x2logxdx+c ....(ii)
 Let I=2logxx2dx
I=2[logxx2dx{1xx2dx}dx]I=2[logxx111xx11dx]
=2(logx1x+1x2dx)I=2(1xlogx1x)=2x(1+logx)
Now enq (ii) becomes
ylogx=I+Cylogx=2x(1+logx)+c

Differential Equation Exercise Revision Exercise (RE) Question 66 (xii)

Answer : y=(1+x2)1log|sinx|+c(1+x2)1
Hint : integrate by applying integration of xn
Given : (1+x2)dy+2xydx=cotxdx
Solution : given equation
(1+x2)dy+2xydx=cotxdx
Divide both sides by dx
(1+x2)dydx+2xydxdx=cotxdxdx(1+x2)dydx+2xy=cotx
Divide by (1+x2)
dydx+2xy1+x2=cotx1+x2
Comparing above equation with dydx+Py=Q
P=2x1+x2,Q=cotx1+x2 I. F=ePdx I. F=e2x1+x2dx Let t=1+x2
dt=2xdxI.F=edtt=elogt=t=(1+x2)
y×I.F=Q×I.Fdx+cy(1+x2)=cotx1+x2×(1+x2)dx+cy(1+x2)=cotxdx+c
y(1+x2)=log|sinx|+c Divide 1+x2y=(1+x2)1log|sinx|+c(1+x2)1

Differential Equation Exercise Revision Exercise (RE) Question 66 (xiii)

Answer : x+y+1=cey
Hint : you integrate by applying integration of xn
Given : (x+y)dydx=1
Solution : Put in the form of dydx+Py=Q
(x+y)dydx=1
Divide by x+y
dydx=1x+y
dxdy=x+y
dxdy+(x)=y .....(i)
Find P1 and Q1
Comparing (i)
dydx+P1=Q1P1=1,Q1=y
Find I.F
 I.F =eP1dy
=e1dy=ey
x×I.F=Q1×I.Fdy+c
Putting value xey=y×eydy+c ......(ii)
 Let I=yeydy
=yeydy(ddyyeydy)dy
=yey1ey1dy
=yey+eydy
I=yeyey
Put value of I in (ii)
xey=yeydy+cxey=yeyey+c
Divide by ey
x=y1+ceyx+y+1=cey

Differential Equation Exercise Revision Exercise (RE) Question 66 (xiv)

Answer : x=y33+cy
Hint : integrate by applying integration of xn
Given : ydx+(xy2)dy=0
Solution : ydx+(xy2)dy=0
dydx=yxy2
This is not in the form of dydx+Py=Q
dxdy=y2xy
dxdy=yxy
dxdy+xy=y .....(i)
Find P and Q Where P=1y,Q=y
Find I.F
 I.F =ePdy=e1ydy=elogy=y
Solution will be
x(I.F)=(Q×I.F)dy+cxy=y×ydy+cxy=y2dy+c
xy=y33+cx=y23+cy

Differential Equation Exercise Revision Exercise (RE) Question 66 (xv)

Answer : x=3y2+cy

Hint : integrate by applying integration of xn

Given : (x+3y2)dydx=y

Solution : (x+3y2)dydx=y

This is not in the foem of dxdy+Px=Q

Where P1=1y&Q=3y
Step : 3 find integration factor
 I. F=ePdy I. F=e1ydyelogy=elogy1=y1=1y
Step : 4
Solution is
x(I.F)=Q×I.Fdy+cx1y=3y1ydy+cxy=3dy+c
xy=3y+c
x=3y2+cy


Differential Equation Exercise Revision Exercise (RE) Question 67 (i)

Answer : y(1+x2)=tan1xπ4
Hint : Use the formula 11+x2dx=tan1x+c
Given : (1+x2)dydx+2xy=11+x2;y=0 when x=1
Solution : (1+x2)dydx+2xy=11+x2
Divide by 1+x2
dydx+(2x1+x2)y=1(1+x2)2
This is a linear differential equation of the form
Here, P=2x1+x2;Q=1(1+x2)2
The integrating factor of this differential equation is
 I.F =ePdx=e2x1+x2dx
We have, 2x1+x2=log|1+x2|+c[dxx=log|x|+c]
Therefore,  I. F=elog|1+x2|=1+x2
 I. F=1+x2
Hence the solution of the differential equation is
y(I.F)=(Q×I.F)dx+cy(1+x2)=1(1+x2)2(1+x2)dx+cy(1+x2)=11+x2dx+c ...(i)
We know , 11+x2dx=tan1x+c
Therefore, y(1+x2)=tan1x+c ....(ii)
Now y=0 when x=1
Therefore,
0(1+12)=tan1(1)+c0=π4+cc=π4
By (ii)
y(1+x2)=tan1xπ4


Differential Equation Exercise Revision Exercise (RE) Question 67 (ii)

Answer : tan1yx+log|y2x2+1×x|=π4+12log2
Hint : using variable separable method and substituting the values
Given : (x+y)dy+(xy)dx=0,y=1 when x=1
Solution :
(x+y)dy+(xy)dx=0(x+y)dy=(xy)dx(x+y)dy=(yx)dxdydx=yxx+y=(xy)x+y
Now let ,
f(x,y)=dydx=(xy)x+y Finding f(λx,λy)
f(λx,λy)=(λxλy)λx+λy=λ(xy)λ(x+y)=(xy)x+y=λ0f(x,y)
Therefore, f(x,y) is a homogeneous function of degree 0 .
Putting y=vx
Diff w.r.t.x
dydx=xdvdx+v Putting value of dydx and y=vx in (i) 
dydx=(xy)x+yv+xdvdx=(xvx)x+vxv+xdvdx=x(1v)x(1+v)
=v11+vxdvdx=v11+vvxdvdx=v1v(1+v)1+v
xdvdx=(1+v2)1+v1+v1+v2dv=dxx
Integrating both sides
1+v1+v2dv=dxx11+v2dv+vv2+1dv=log|x|+ctan1v+vv2+1dv=log|x|+c
Putting v2+1=t in integral
2vdv=dtvdv=dt2
tan1v+1t×dt2=log|x|+ctan1v+12log|t|=log|x|+ctan1v+12log|v2+1|=log|x|+c
Now again putting back the value of v=yx
tan1yx+12log|(yx)2+1|+log|x|=ctan1yx+log|(yx)2+1|+log|x|=ctan1yx+log|(yx)2+1×x|=c ....(ii)
Now, y=1 when x=1
 Therefore tan1(11)+log|(11)2+1×1|=ctan1(1)+log2=cπ4+12log2=c
Put in (ii)
tan1yx+log|y2x2+1×x|=π4+12log2

Differential Equation Exercise Revision Exercise (RE) Question 67 (iii)

Answer : y+2x=3x2y
Hint : using variable separable method and substituting the values
Given : x2dy+(xy+y2)dx=0,y=1, when x=1
Solution : x2dy+(xy+y2)dx=0
The differential equation can be written as
x2dy=(xy+y2)dxdydx=(xy+y2)x2 ......(i)
Let f(x,y)=dydx=(xy+y2)x2
Finding f(λx,λy)
f(λx,λy)=(λx.λy+λ2y2)λ2x2=λ2(xy+y2)λ2x2=(xy+y2)x2=λ0(f(x,y))f(x,y)=(xy+y2)x2
Therefore , f(x,y) is a homogenous function of degree zero.
Putting y=vx
Diff w.r.t.x
dydx=xdvdx+v
Putting value of dydx and y=vx in (i)
dydx=(xy+y2)x2v+xdvdx=(x(vx)+(vx)2)x2v+xdvdx=(x2v+x2v2)x2v+xdvdx=x2(v+v2)x2
xdvdx=vv2v=(v2+2v)dvv2+2v=dxx
Integrating both sides we get ,
dvv2+2v=dxx
dv(v2+2v+1)1=logx+logc
dv(v+1)212=logx+logc
Using dxx2a2=12alog|xax+a|+c
12logv+11v+1+1=logx+logc
12logvv+2=logx+logc
logvv+2+logx=logcxvv+2=c
Now putting value of v i.e , y/x
xyxyx+2=c
x2×yxyx+2=c
xyy+2xx=c
xyy+2x=c
xy=cy+2x
Squaring both sides
x2y=c2(y+2x)(ii) Now x=1,y=1 Therefore (12)(1)=c2(1+2(1))
1=c2(3)c2=13
Putting back in (ii)
x2y=13(y+2x)3x2y=y+2xy+2x=3x2y

Differential Equation Exercise Revision Exercise (RE) Question 68

Answer : y=x2+log|x|
Hint : Using variable separable method
Given : xdy=(2x2+1)dx,x0
Solution : xdy=(2x2+1)dx
dy=(2x2+1)xdxdy=(2x2x+1x)dxdy=(2x+1x)dx
Integrating both sides
dy=(2x+1x)dx ....(i)
dy=2xdx+1xdx
y=2x22+log|x|+c
y=x2+log|x|+c .....(ii)
Since the curve passes through (1,1)
Putting x=1,y=1 in (ii)
1=(1)2+log(1)+c[log1=0]1=1+0+c=>c=0
Put c=0 in (ii)
i.e , y=x2+log|x|+c
y=x2+log|x|+0
y=x2+log|x|
Hence the equation of curve is
y=x2+log|x|

Differential Equation Exercise Revision Exercise (RE) Question 69

Answer : y=(3x2+15)13
Hint : use variable separable method
Given : slope of the tangent to the curve at any (x,y) is 2xy2
Solution : Slope of tangent =dydx
dydx=2xy2y2dy=2xdx
Integrating both sides
y2dy=2xdx
y33=2x22+c
y33=x2+c
y3=3x2+3c
y3=3x2+c1..... (i) where c1=3c
Given that equation passes through (-2, 3)
Putting x=2,y=3 in (i)
y3=3x2+c133=3(2)2+c127=3×4+c1c1=2712=15
Putting C1 in (i)
y3=3x2+15y=(3x2+15)13 is the perticular solution of the equation

Differential Equation Exercise Revision Exercise (RE) Question 70

Answer : 2y1=ex(sinxcosx)
Hint : using integration by parts
Given : dydx=exsinx
Solution : dydx=exsinx
dy=exsinxdx
Integrating both sides
dy=exsinxdxy=exsinxdx ....(i)
Using integration by parts
Let I=exsinxdx=sinxexcosxexdx
=sinxex[cosxex(sinx)exdx]
I=sinxexcosxexsinxexdx
I=sinxexcosxexI
2I=sinxexcosxex
I=ex(sinxcosx)2
Put in (i), we get,
y=ex(sinxcosx)22y=ex(sinxcosx)y=12ex(sinxcosx)+c ....(ii)

Given curve passes through (0,0)

Putting x = 0 , y = 0 in equation

Therefore , 0=12e0(sin0cos0)+c

0=12(1)(01)+c[e0=1]

0=12+c

c=12

Putting value of C in (ii)

y=12ex(sinxcosx)+12y12=12ex(sinxcosx)2y12=ex2(sinxcosx)2y1=ex(sinxcosx)


Differential Equation Exercise Revision Exercise (RE) Question 71

Answer: y+3=(x+4)2
Given:
At any point (x,y)of a curve , the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (4,3)
Hint:
Using variable separable method and substituting the values.
Explanation:
Slope of the tangent to the curve =dydx
Slope of line segment joining (x,y) and (4,3)
=y2y14+x
=3y4x
=(y+3)(x+4
=y+3x+4
Given at point (x,y) . Slope of tangent is twice of line segment
dydx=2(y+3x+4)2dyy+3=2dxx+4
Integrating both sides.
dyy+3=2dxx+4
log|y+3|=2log|x+4|+logc
log|y+3|=log|x+4|2+logc[alogx=logx2]
log(y+3)log(x+4)2=logc
log(y+3)(x+4)2=logcy+3(x+4)2=c(1)

Since curve passes through (-2,1)
Put x=2andy=1 in (1)

1+3(2+4)2=C

C=4(2)2

=44=1C=1

Put value of c in eq (1)

y+3(x+4)2=1y+3=(x+4)2
Hence the equation of the curve is
y+3=(x+4)2


Differential Equation Exercise Revision Exercise (RE) Question 72

Answer :x2y2=cx
Given : dydx=x2+y22xy
Hint:
Using variable separable method.
Explanation:
We know that the slope of the tangent at(x,y) of a Curve is dydx

Given slope of tangent at (x,y) is x2+y22xy
dydx=x2+y22xy(1)
 Put dydx=f(x,y)F(x,y)=x2+y22xy Finding f(λx,λy)=(λx)2+(λy)22(λx)(λy)
=λ2x2+λ2y22λ2xy
=λ2(x2+y2)λ22xy
=x2+y22xy
=F(x,y)
 So, F(λx,λy)=F(x,y)
=λF(x,y)
So F(λx,λy) is homogeneous function of degree 0
Given equation is homogeneous differential equation
Now Put y=vx in (1)
Differentiate with respect to x
dydx=xdvdx+v Putting dydx value in (1) and y=vx
xdvdx+v=x2+(vx)22x(vx)=x2+v2x22vx2=1+v22vxdvdx=1+v22vvxdvdx=1+v22v22v
xdvdx=1v22vdvdx=1v22v1x2v1v2dv=dxx
2v21dv=dxx2vdvv21=dxx
Integrating both sides,
2vv21dv=dxxvv21 dv=dxxvv21dv=log|x|+c
Solving 2vv21dv
Put v21=t
Differentiate with respect to v
2vdv=dtdv=dt2v=2vtdt2v=dtt=log1+1
Putting back
t=v21log|v21|
By (2)
log|v21|=log|x|+c1 Putting vx=y or 
v=yxlog|(yx)21|=log|x|+C1log|(yx)21|+log|x|=C1
log|[(yx)21]x˙|=C1log|[y2x21]x|=C1
Putting C1=logC1
log|(y2x21)x|=logC1
Remove log of both sides.
(y2x21)x=C1
xy2x2x=C1
y2xx=C1
y2x2=xC1x2y2=C1x Put C=C1x2y2=cx
Hence Proved


Differential Equation Exercise Revision Exercise (RE) Question 73

Answer: 1+2ex2/2
Given: The slope of the tangent to the curve at any point (x,y) is equal to the sum of x coordinate and the product of x and y coordinate of that point
Hint: You must know about integrating factor
Explanation: Slope of the tangent to the curve at (x,y)=dydx
Given that,
Slope of the tangent to the curve at point (x,y) is equal to the sum of x coordinate and the product of x and y coordinate of that point
So our equation becomes,
dydx=x+xydydxxy=x
Differential equation is of the form
dydx+Py=Q
Where
P=x or θ=x
If
=ePdx=exdx=ex2/2
Solution is
y(If)=(θ×If)dx+cyex22=xex22dx+c
Putting ,
x22=t2x2dx=dtxdx=ab
Thus our equation becomes
yex2/2=etdt+cyex2/2=et+c Putting back t=x22yex22=ex/2+cyex2/2
=1ex2/2+cy=1+cex2/2(1)
Since curve passes through (0,1)
Putting
x=0,y=1in(1)y=1+cex2/21=1+cea2/21+1=C [e0=1]
c=2
Putting value of c in (1)
y=1+cex2/2y=1+2ex2/2
Equation of the curve is 1+2ex2/2

Differential Equation Exercise Revision Exercise (RE) Question 74

Answer : x+y+1=ex
Given: The slope of the tangent to the curve at any point (x,y) is equal to the sum of x coordinate and the product of x and y coordinate of that point
Hint: You must know about integrating factor
Explanation: We know that
Slope of the tangent to the curve at (x,y)=dydx
According to the given
dydx=x+xydydxxy=x
This is of the form
dydx+Py=Q
Where
P=1 and θ=x
Finding integrating factor
If,
=ePdx=e(1)dx=exdx [1dx=x+c]
Solution is
y(If)=(θ×If)dx+cyex=xexdx+c
Using integration by parts,
xexdx=xex1(1)ex1dx=xex+ex1+c=xexex+C
Put in (1)
yex=xexex+c
Divide by ex
y=x1+cex(2)
Since curve passes through origin,
Putting x=0 and y=0 in (2)
0=01+Ce0c=1 [e0=1]
Put vaue of c in (2)
y=x+1+exx+y+1=ex

Differential Equation Exercise Revision Exercise (RE) Question 75

Answer: y={4x2ex iff dydx=+(x+y5)6x4ex iff dydx=(x+45)
Given: Sum of the coordinate of any point of curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5
Hint: Using integration by parts and integration factor
Explanation: We know that
Slope of the tangent to the curve at (x,y)=dydx
According to the given
|dydx|+5=x+y|dydx|=x+y5dydx=±(x+y5)
So we will take both (+) ve and (-) ve sign and then solve it
Taking (+) ve sign:
dydx=x+y5dydxy=x5
Equation is of the form
dydx+Py=Q
Where,
P=1θ=x5IF=ePdx=e(1)dx=ex
Solution is
y(IF)=(θ×If)dx+cyex=(x5)exdx+cyex=(x5)ex(1)ex1dx+cyex=(x5)ex+exdx+cyex=(x5)exex+c
Divided by ex
y=(x5)1+cex(1)yy=5x1+cex=4x+cex
Since the curve passes through -the point (0,2)
Put x=0 and
y=2
2=40+ce2=4+c[e0=1]C=2
Equation of the curve is
y=4x2ex
Taking (-) ve sign:
dydx=xy+5dydx+y=x+5
Equation is of the form
dydx+Py=Q
Where,
P=1θ=x+5IF=ePdx=e(1)dx=ex [1dx=x+c]
Solution is
y(IF)=(θ×If)dx+cyex=(5x)ex(1)exdx+c [Using integration by parts]
yex=(5x)ex+ex+c
Divided by ex
yex=(x5)exex+cy=(x5)1+cex(1)yy=5x1+cex=4x+cex
Since the curve passes through the point (0,2)
Put x=0 and
y=22=60+ce2=6+c(1) [e0=1]
C=4
Put in (2) we get,
Equation of the curve is
y=6x4ex

Differential Equation Exercise Revision Exercise (RE) Question 76

Answer : y2=x+5
Given: The slope of the tangent to the curve at any point is the reciprocal of twice the ordinate at that point. Also the curve passes through the point(4,3)
Hint: Using variable separable method
Explanation: Let P(x,y) be any point on the curve
Slop of the tangent at P(x,y)=dydx
Acc to given condition,
dydx=12y2ydy=1dx
Integrating both sides,
2ydy=1dx2y22=x+cy2=x+c
Since the curve passes through (4,3) then,
a=4+cC=5
Hence the required equation of the curve is
y2=x+5

Differential Equation Exercise Revision Exercise (RE) Question 77

Answer : log2λ
Given: The decay rate of radium at any time t is proportional to its mass at that time.
Hint: Using variable separable method
Explanation: Acc to given,
dmdtαmdmdt=dm
 where λ>0 [-ve sign  of decay rate ]dmm=dt
Integrating both sides
dmm=λdtlogm=λ+c(1) Let m0 be the initial mass at t=0logm0=λ0+cc=logm0
Put in (1)
logm=λt+logm0logmlogm0=λtlog(mm0)=λt2 [logmlogx=logmh]
We have to find the time when m=12m0
2m=m0
Put in (2)
logm2 m=λtlog12=λtlog1log2=λt0log2=λt [log1=0]
λt=log2t=log2λ

Differential Equation Exercise Revision Exercise (RE) Question 78

Answer: 0.04%
Given: Radius disintegrates at a rate proportional to the amount of radium present at the moment
Hint: Using variable separable method
Explanation: Let A be the amount radium at that time
According to given,
dAdtAdAdt=λA,λ>0dAA=λdt
Integrating both sides,
dAA=λdtlogeA=λt+c(1)
Let A0 be the initial amount at t=0
logeA0=ClogeA=λt+logeA0logeAlogeA0=λtlogeAA0=λtlogA0 A=λt
Now,
A=Ao2t=1590λ×1590=log2 A A
log21590=λλ=logc(1+0A)t=1log21590=logc
A0AA0A=elog21590A=elog21590A0A=0.9996A0 [of given]
 Reqd. %=AA0 A0×100=0.9996 A0A0 A0×100=0.04%

Differential Equation Exercise Revision Exercise (RE) Question 79

Answer : T=log20log2
Given: A wet porous substance in the wet air loses its moisture at a rate proportional to the moisture content
Hint: Using variable separable method
Explanation: Let M be the moisture content at any time A
According to given,dMdtαMdmdt=km

when k be any constant and [-ve sign. ? of it losses its moisture]

dmm=kdt

Integrating both the sides.

dmm=kdtlogm=kt+c(1) Let at t=0M=M0

Put in (1)

logM=kt+logM0logMlogM0=Ktlog|MM0|=lct(2) [logalogb=logab]

Given tht at t=1,

m=12×m0

Put in (2)

log|M2M|=k×1log12=Klog21=klog21=k [logab= blo g a]

1log2=kk=log2

Put in (2)

log|MM0|=log2tlog|M0M|=log2t(3)

Let at t=T,

moisture = 95%

Remaining = 5 %

M=5% of M0M=5100×M0=120M0

Put in (3)

log(20MM)=+log2Tlog20=log2TT=log20log2

Class 12 maths book consists of eleven exercises from the 21st chapter. There are 113 revision-based exercises in this material. The concepts on which these questions are framed are, differentiating various sums, integration factor of differential equations, formation of differential equations, solving first-order differential equations, and many more. It may seem to be quite enormous, but all it takes is a good practice with the RD Sharma Class 12th Exercise solution book. Therefore, the importance of the RD Sharma Class 12 Solutions Differential Equations Ex RE book is ineffable.

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RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

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