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RD Sharma Class 12 Exercise 21.2 Differential Equation Solutions Maths-Download PDF Online

RD Sharma Class 12 Exercise 21.2 Differential Equation Solutions Maths-Download PDF Online

Edited By Satyajeet Kumar | Updated on Jan 24, 2022 04:19 PM IST

Class 12 students face regular tests and exams to test their understanding ability of almost every concept they learn. This requires regular practice in the right manner. Topics like Differential equations in mathematics are always challenging for high school students. Students must have the RD Sharma Class 12th Exercise 21.2 to have clarity in this chapter and attend those sums without any doubts. Not only do students use these solution books, many teachers and tutors also do to recheck their answers and find easier methods to solve the sums.

This Story also Contains
  1. RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise
  2. Differential Equations Excercise: 21.2
  3. RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

Differential Equations Excercise: 21.2

Differential Equation exercise 21.2 question 1

Answer:
8(dydx)327y=0
Hint:
Differentiating the given equation
Given:
y2=(xc)3
Solution:
y2=(xc)3
2ydydx=3(xc)2
2ydydx3=(xc)
y2=(xc)3=(2ydydx3)32
y4=(2ydydx3)3
y4=8y327(dydx)3
y=827(dydx)3
8(dydx)327y=0




Differential Equation exercise 21.2 question 2

Answer:
xdydx=ylogy
Hint:
Differentiating given equation with respect to x
y=emx [Applying log both sides]
Given:
y=emx
Solution:
dydx=memx
dydx=my
We have,
y=emx
logy=mx
m=logyx
dydx=ylogyx
xdydx=ylogy
Hence,
xdydx=ylogy
is the differential equation corresponding to y=emx


Differential Equation exercise 21.2 question 3 (i)

Answer:
2x(dydx)=y
Hint:
Differentiating the given equation with respect to x
Given:
y2=4ax
Solution:
y2=4ax
a=y24x
On differentiating with respect to x,
2ydydx=4a
Substituting the value of a, we get
2ydydx=4y24x2ydydx=y2x2x(dydx)=y
Hence,
2x(dydx)=y
is the differential equation corresponding to y2=4ax


Differential Equation exercise 21.2 question 3 (ii)

Answer:
(dydx)3+2(dydx)2+xdydxy=0
Hint:
Differentiating the given curve, c- Parameter
Given:
y=cx+2c2+c3
Solution:
y=cx+2c2+c3
dydx=c+0c=dydxy=xdydx+2(dydx)2+(dydx)3
(dydx)3+2(dydx)2+xdydxy=0


Differential Equation exercise 21.2 question 3 (iii)

Answer:
y+xdydx=0
Hint:
Differentiating the given equation, where a-Parameter
Given:
xy=a2
Solution:
xy=a2
ddx(xy)=ddx(a2)(1)y+xdydx=0y+xdydx=0



Differential Equation exercise 21.2 question 3 (iv)

Answer:
d3ydx3=0
Hint:
Differentiating the given equation
Given:
y=ax2+bx+c
Solution:
y=ax2+bx+c
dydx=ddx(ax2+bx+c)dydx=2ax+bd2ydx2=ddx(2ax+b)
d2ydx2=2ad3ydx3=ddx(2a)d3ydx3=0


Differential Equation exercise 21.2 question 4

Answer:
d2ydx24y=0
Hint:
Differentiating the given equation in first and second order
Given:
y=Ae2x+Be2x
Solution:
y=Ae2x+Be2x
dydx=2Ae2x2Be2xd2ydx2=4Ae2x+4Be2xd2ydx2=4yd2ydx24y=0


Differential Equation exercise 21.2 question 5

Answer:
d2xdt2+n2x=0
Hint:
Differentiating the given equation with respect to x.
Let the same equation and substitutes its value.
Given:
x=Acosnt+Bsinnt
Solution:
x=Acosnt+Bsinnt
dxdt=Ansin(nt)+Bncos(nt)d2xdt2=An2cos(nt)+Bn2sin(nt)d2xdt2=n2[Acos(nt)+Bsin(nt)]
We get,
d2xdt2=n2xd2xdt2+n2x=0


Differential Equation exercise 21.2 question 6

Answer:
xyd2xdt2+x(dydx)2ydydx=0
Hint:
Differentiating the given equation, let it equation (i) and substitute a from (iii) in (ii)
Given:
y2=a(bx2)
Solution:
y2=a(bx2)
2ydydx=a(2x)ydydx=ax(ii)
Again,
ddx(ydydx)=ddx(ax)(dydx)2+yd2ydx2=a(iii)
Substitute equation (iii) in (ii),
ydydx=[(dydx)2+yd2ydx2]xydydx=x(dydx)2+xyd2ydx2xyd2ydx2+x(dydx)2ydydx=0


Differential Equation exercise 21.2 question 7

Answer:
y2(x22y2)4xyyx2=0
Hint:
Differentiating the given equation with respect to x
Given:
y22ay+x2=a2
Solution:
y22ay+x2=a2
Differentiating with respect to x
2ydydx2adydx+2x=0ydydx+x=adydxa=ydydx+x(dydx)
Putting the value of a in the given equation, we get
y22a(x+ydydxdydx)y+x2=(x+ydydxdydx)2
put
dydx=y
y22a(x+yyy)y+x2=(x+yyy)2y2y2(y2y+xy)+x2yy=y2y2+2xyy+x2y2
y2y22y2y22xyy+x2y2y2y2xyyx2=04xyy+x2y2x22y2y2=0y2(x22y2)4xyyx2=0


Differential Equation exercise 21.2 question 8

Answer:
[1+(dydx)2]3=r2(d2ydx2)2
Hint:
Differentiating the given equation where r is a constant
Given:
(xa)2+(yb)2=r2(i)
Solution:
(xa)2+(yb)2=r2
2(xa)+2(yb)dydx=0(ii)1+(yb)d2ydx2+(dydx)2=0(yb)d2ydx2=(1+(dydx)2)(iii)
From equation (ii)
(xa)=(yb)dydx
Put this value in equation (i)
[(yb)dydx]2+(yb)2=r2(iv)
Squaring equation (iii),
(yb)2(d2ydx2)2=(1+(dydx)2)2(v)
Dividing equation (iv) by (v),
(dydx)2+1(d2ydx2)2=r2[(dydx)2+1]
[1+(dydx)2]3=r2(d2ydx2)2


Differential Equation exercise 21.2 question 9

Answer:
(x2y2)dydx2xy=0
Hint:
Let (0, k) be the centre of the circle with k as its centre and differentiating the equation of the circle
Given:
Circles pass through origin and their centre lie on y-axis
Solution:
(x0)2+(yk)2=k2x2+(yk)2=k2x2+y22ky=0x2+y22y=k
2y(2x+2ydydx)(x2y2)2dydx4y2=04y(x+ydydx)2(x2+y2)dydx=04xy=4y2dydx2(x2+y2)dydx=0
(4y22x22y2)dydx+4xy=0(2y22x2)dydx+2xy=0(y2x2)dydx+2xy=0(x2y2)dydx2xy=0


Differential Equation exercise 21.2 question 10

Answer:
x2y2+2xydydx=0
Hint:
Let the equation of the circle assuming R as radius and (a,b) as the centre of the circle is
(xa)2+(yb)2=R2
Given:
Circles passes through origin and their centre lie on x-axis. So the centre of the circle will be (a, 0)
(xa)2+y2=a2(i)
Solution:
(xa)2+y2=a22(xa)dx+2ydy=0xa=ydydxa=x+ydydx
Substituting these values in equation (i)
x2+y2=2x(x+ydydx)x2+y2=2x2+2xydydxy2x22xydydx=0x2y2+2xydydx=0


Differential Equation exercise 21.2 question 11

Answer:
drdt=k
Hint:
Let the surface be A
Rate of evaporation,
dvdt
Using formula,
v=43πr3,A=4πr2
Given:
Rain drop evaporates at a rate proportional to its surface area
Solution:
dvdtAdvdt=kAddt(43πr3)=k(4πr2)43π(3r2)drdt=4πr2drdt=k



Differential Equation exercise 21.2 question 12

Answer:
2ad2ydx2+(dydx)3=0
Hint:
Let the equation of parabola be
(yk)2=4a(xh)
Where h, k are parameter
Given:
Latus rectum 4a and axis are parallel to x-axis
Solution:
Let the equation of parabola be
(yk)2=4a(xh)
Where h, k are parameter
2(yk)dydx=4a(yk)dydx=2a(yk)d2ydx2+(dydx)2=02ad2ydx2+(dydx)3=0
which is the required differential equation.



Differential Equation exercise 21.2 question 13

Answer:
dydx+2xy=4x3
Hint:
y=2(x21)+cex2y=2x22+cex22x2y=2cex2 Substitutes 2cex2 by 2x2y
Given:
dydx=4x+cex2(2x)dydx=2x(2cex2)dydx=2x(2x2y)dydx=4x32xydydx+2xy=4x3
Hence proved.


Differential Equation exercise 21.2 question 14

Answer:
d2ydx2(1x2)dydxx2=0
Hint:
Differentiating the given equation with respect to x
Given:
y=(sin1x)2+Acos1x+B
Solution:
dydx=2sin1x11x2+A(11x2)+0y=11x2[2sin1xA]y1x2=2sin1xAA=2sin1xy21x2
0=21x2y1x2y21(2x)21x20=21x2y1x2+y2x1x20=2y(1x2)+y3x1x2
y(1x2)yx2=0
d2ydx2(1x2)dydxx2=0


Differential Equation exercise 21.2 question 15 (i)

Answer:
2xydydx+4x2y2=0
Hint:
Using chain rule and putting the value of a in (i)
Given:
(2x+a)2+y2=a2(i)
Solution:
2(2x+a)2+2ydydx=02(2x+a)+ydydx=0(2x+a)=12ydydxa=12ydydx2x
(12ydydx)2+y2=(y2dydx+2x)2(12ydydx)2+y2=(y2dydx)2+4x2+2y2.dydx.2xy24x22xydydx=0
2xydydx+4x2y2=0


Differential Equation exercise 21.2 question 15 (ii)

Answer:
2xydydx(y2+4x2)=0
Hint:
Differentiating the given equation
Given:
(2xa)2y2=a2
Solution:
2(2xa)22ydydx=02xa=y2dydxa=(2xx2dydx)(y2.dydx)2y2=4x2+(y2.dydx)22.2a.y2.dydx
2xydydx(y2+4x2)=0


Differential Equation exercise 21.2 question 15.3 (iii)

Answer:
4xydydx+x22y2=0
Hint:
Using the formula (a+b)2 and differentiating the given equation
Given:
(xa)2+2y2=a2
Solution:
2(xa)+2y.2dydx=0(xa)=2ydydxa=x+2ydydx(2+dydx)2+2y2=x2+(2+dydx)2+2x.2ydydx
4xydydx+x22y2=0


Differential Equation exercise 21.2 question 16 (i)

Answer:
The required equation is
x+ydydx=0
Hint:
The given equation is the equation of a circle
Given:
x2+y2=a2
Solution:
x2+y2=a2
Differentiating with respect to x,
2x+2ydydx=0x+ydydx=0
The required equation is
x+ydydx=0


Differential Equation exercise 21.2 question 16 (ii)

Answer:
The required equation is
xydydx=0
Hint:
Differentiating the given equation with respect to x
Given:
x2y2=a2
Solution:
x2y2=a2
Differentiating with respect to x,
2x2ydydx=0xydydx=0
The required equation is
xydydx=0


Differential Equation exercise 21.2 question 16 (iii)

Answer:
The required equation is
y2xdydx=0
Hint:
The equation is the equation of parabola
Given:
y2=4ax
Solution:
y2=4ax
2ydydx=4adydx=2aydydx=2y(y24x)dydx=y2x2xdydx=y
The required equation is
y2xdydx=0


Differential Equation exercise 21.2 question 16 (iv)

Answer:
The required differential equation is
x2[1+(dydx)2]=(dydx)2
Hint:
 Putting 1x2 instead of (yb) in 2x+2(yb)dydx=0
Given:
x2+(yb)2=1
Solution:
The equation of family of curves is
x2+(yb)2=1(i)
Where b is a parameter
Differentiating equation (i) with respect to x, we get
2x+2(yb)dydx=02x+21x2dydx=0[Using(i)]x=1x2dydxx2=(1x2)(dydx)2x2=(dydx)2x2(dydx)2
The required equation is
x2[1+(dydx)2]=(dydx)2




Differential Equation exercise 21.2 question 16 (v)

Answer:
The required differential equation is
y2(dydx)2y2=1
Hint:
Differentiating equation (i) with respect to x
Given:
(xa)2y2=1
Solution:
The equation of family of curves is
(xa)2y2=1(i)
Where a is a parameter
Differentiating equation (i) with respect to x, we get
2(xa)2ydydx=0(xa)ydydx=01+y2=ydydx[Using(i)](1+y2)=y2(dydx)2
The required equation is
y2(dydx)2y2=1


Differential Equation exercise 21.2 question 16 (vi)

Answer:
The required differential equation is
x[yd2ydx2+(dydx)2]=ydydx
Hint:
This is the equation of hyperbola
x2a2y2b2=1
Given:
x2a2y2b2=1
Solution:
The equation of family of curves is
x2a2y2b2=1(i)
Where a and b are parameter
Differentiating equation (i) with respect to x, we get
2xa22yb2dydx=0(ii)
Differentiating equation (ii) with respect to x, we get
2a22b2(dydx)22yb2(d2ydx2)=02a2=2b2[y(d2ydx2)+(dydx)2]b2a2=[y(d2ydx2)+(dydx)2](iii)
Now, from (ii), we get
2xa2=2yb2dydxb2a2=yxdydx(iv)
From (iii) and (iv), we get
yxdydx=[y(d2ydx2)+(dydx)2]
The required differential equation is
x[yd2ydx2+(dydx)2]=ydydx


Differential Equation exercise 21.2 question 16 (vii)

Answer:
The required differential equation is
yd2ydx2+(dydx)2=0
Hint:
Differentiating equation (i) and (ii) with respect to x
Given:
y2=4a(xb)
Solution:
The equation of family of curves is
y2=4a(xb)(i)
Where a and b are parameters
Differentiating equation (i) with respect to x, we get
2ydydx=4aydydx=2a(ii)
Differentiating equation (ii) with respect to x, we get
yd2ydx2+(dydx)2=0
The required differential equation is
yd2ydx2+(dydx)2=0


Differential Equation exercise 21.2 question 16 (viii)

Answer:
The required equation is
xdydx=3y
Hint:
Differentiating the given equation with respect to x
Given:
y=ax3
Solution:
y=ax3
On differentiating with respect to x, we get
dydx=3ax2
From the given equation,
a=yx3
So, we have
dydx=3yx3(x2)dydx=3yxxdydx=3y
The required equation is
xdydx=3y


Differential Equation exercise 21.2 question 16 (ix)

Answer:
The required equation is
2xydydx=(x2+3y2)
Hint:
Differentiating the given equation with respect to x
Given:
x2+y2=ax3
Solution:
x2+y2=ax3a=x2+y2x3
Differentiating with respect to x,
[(2x+2ydydx)x33x2(x2+y2)]x6=02x4+2x3ydydx3x43x2y2=02x3ydydxx43x2y2=0
2x3ydydx=x4+3x2y22x3ydydx=x2(x2+3y2)2xydydx=(x2+3y2)
The required equation is
2xydydx=(x2+3y2)



Differential Equation exercise 21.2 question 16 (x)

Answer:
The required equation is
xdydx=ylogy
Hint:
Differentiating the given equation with respect to x
Given:
y=eax
Solution:
y=eax
Differentiating with respect to x
dydx=aeaxdydx=ay
From the given equation, we have
y=eaxlogy=axa=logyx
Now,
dydx=ay=ylogyxxdydx=ylogy
The required equation is
xdydx=ylogy


Differential Equation exercise 21.2 question 17

Answer:
The required differential equation is
xyd2ydx2+x(dydx)2ydydx=0
Hint:
Ellipse centre at the origin and foci on the x-axis
Given:
Ellipse having the centre at the origin and foci on the x-axis
Solution:
Equation of required ellipse is
x2a2+y2b2=1(i)
Where a and b are arbitrary constants
Differentiating equation (i) with respect to x, we get
2xa2+2yb2dydx=02xa2=2yb2dydxb2a2=yxdydx(ii)
Differentiating equation (ii) with respect to x, we get
0=yxd2ydx2+(xdydxy)x2dydxxyd2ydx2+x(dydx)2ydydx=0
The required differential equation is
xyd2ydx2+x(dydx)2ydydx=0


Differential Equation exercise 21.2 question 18

Answer:
The required differential equation is
x(dydx)2+xyd2ydx2ydydx=0
Hint:
Hyperbola centre at the origin and foci on the x-axis
Given:
Hyperbolas having the centre at the origin and foci on the x-axis
Solution:
Equation of required ellipse is
x2a2y2b2=1(i)
Where a and b are arbitrary constants
Differentiating equation (i) with respect to x, we get
2xa22yb2dydx=0xa2yb2dydx=0(ii)
Differentiating equation (ii) with respect to x, we get
1a21b2[(dydx)2+yd2ydx2]=01a2=1b2[(dydx)2+yd2ydx2]
Substituting this value of 1/a2 in equation (ii), we get
x[1b2[(dydx)2+yd2ydx2]]yb2dydx=0x[(dydx)2+yd2ydx2]ydydx=0x(dydx)2+xyd2ydx2ydydx=0
The required differential equation is
x(dydx)2+xyd2ydx2ydydx=0



Differential Equation exercise 21.2 question 19

Answer:
The required differential equation is
[(x+y)2+1](dydx)2=(x+ydydx)2
Hint:
The equation belongs to the family of a circle
(xb)2+(yk)2=r2
Given:
The family of circles in the second quadrant and touching the co-ordinate axes
Solution:
Let c denotes the family of circles in the second quadrant and touching the coordinate axes and let(-a,a) be coordinate of the centre of any member of this circle
Now, the equation representing this family of circle is
(x+a)2+(ya)2=a2(i)x2+y2+2ax2ay+a2=0(ii)
Differentiating (ii) with respect to x, we get
2x+2ydydx+2a2adydx=0x+ydydx+aadydx=0x+ydydx=a+adydxa=(x+ydydx)dydx1
Substituting this value of a in (i), we get
[x+(x+ydydx)dydx1]2+[y(x+ydydx)dydx1]2=[(x+ydydx)dydx1]2
[x(dydx1)+(x+ydydx)]2+[y(dydx1)+(x+ydydx)]2=[x+ydydx]2
(x+y)2(dydx)+(x+y)2=[(x+y)2(dydx)]2
[(x+y)2+1](dydx)2=(x+ydydx)2
The required differential equation is
[(x+y)2+1](dydx)2=(x+ydydx)2


Differential Equation exercise 21.2 question 20

Answer:
The required differential equation is
d2ydx2=1y(dydx)2
Hint:
Differentiating the given equation with respect to x
Given:
y=aebx+5
Solution:
y=aebx+5(i)
Where a and b are arbitrary constants
Differentiating with respect to x
dydx=a.bebx+5(ii)
Again, Differentiating with respect to x
d2ydx2=a.b2ebx+5(iii)
Put
b=1ydydx
from (i) and (ii) in equation (iii), we get
d2ydx2=y.1y2(dydx)2
d2ydx2=1y(dydx)2
The required differential equation is
d2ydx2=1y(dydx)2



Differential Equation exercise 21.2 question 21

Answer:
The required differential equation is
y24y1+4y=0
Hint:
Differentiating the given equation with respect to x
Given:
y=e2x(a+bx)
Solution:
y=e2x(a+bx)(i)
Differentiating with respect to x, we get
y1=e2x(0+b)+(a+bx)e2x.2y1=e2xb+(a+bx)e2x.2(ii)
Putting (i) in (ii)
y1=be2x+2yy1=2y=be2x(iii)
Again, Differentiating with respect to x, we get
y22y1=be2x.2(iv)
Putting (iii) in (iv), we get
y22y1=2y14y
The required differential equation is
y24y1+4y=0


Chapter 21, Differential Equation of class 12, is a portion where most of the students lose their marks. This chapter consists of eleven exercises, ex 21.1 to ex 21.11. this seconds exercise, ex 21.2, contains 27 sums to be solved. The concepts involved in this exercise are solving the differential equations in various forms. Even though the entire chapter has the same concept, the method by which each sum is solved varies greatly. Therefore, the use of RD Sharma Class 12 Chapter 21 Exercise 21.2 becomes essential to make the students understand each method without being confused by one another.

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RD Sharma Chapter-wise Solutions

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