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

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.

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RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise
Differential Equations Excercise: 21.2
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:
$8(\frac{\mathrm{dy} }{\mathrm{d} x})^{3}-27y=0$
Hint:
Differentiating the given equation
Given:
$y^{2}=(x-c)^{3}$
Solution:
$y^{2}=(x-c)^{3}$
$2y\frac{\mathrm{dy} }{\mathrm{d} x}=3(x-c)^{2}$
$\sqrt{\frac{2y\frac{\mathrm{dy} }{\mathrm{d} x}}{3}}=(x-c)$
$y^{2}=(x-c)^{3}=(\frac{2y\frac{\mathrm{dy} }{\mathrm{d} x}}{3})^{\frac{3}{2}}$
$y^{4}=(\frac{2y\frac{\mathrm{dy} }{\mathrm{d} x}}{3})^{3}$
$y^{4}=\frac{8y^{3}}{27}(\frac{\mathrm{dy} }{\mathrm{d} x})^{3}$
$y=\frac{8}{27}(\frac{\mathrm{dy} }{\mathrm{d} x})^{3}$
$8(\frac{\mathrm{dy} }{\mathrm{d} x})^{3}-27y=0$

Differential Equation exercise 21.2 question 2

Answer:
$x\frac{\mathrm{dy} }{\mathrm{d} x}=ylog\: y$
Hint:
Differentiating given equation with respect to x
$y = e^{mx}$ [Applying log both sides]
Given:
$y = e^{mx}$
Solution:
$\frac{\mathrm{dy} }{\mathrm{d} x}=me^{mx}$
$\frac{\mathrm{dy} }{\mathrm{d} x}=my$
We have,
$y = e^{mx}$
$log\: y=mx$
$m=\frac{log\: y}{x}$
$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{y \: log\: y}{x}$
$x\frac{\mathrm{dy} }{\mathrm{d} x}={y \: log\: y}$
Hence,
$x\frac{\mathrm{dy} }{\mathrm{d} x}={y \: log\: y}$
is the differential equation corresponding to $y = e^{mx}$

Differential Equation exercise 21.2 question 3 (i)

Answer:
$2x(\frac{\mathrm{d} y}{\mathrm{d} x})=y$
Hint:
Differentiating the given equation with respect to x
Given:
$y^{2}=4ax$
Solution:
$y^{2}=4ax$
$a=\frac{y^{2}}{4x}$
On differentiating with respect to x,
$2y\frac{\mathrm{d} y}{\mathrm{d} x}=4a$
Substituting the value of a, we get
$\begin{aligned} &2y\frac{\mathrm{d} y}{\mathrm{d} x}=4\frac{y^{2}}{4x} \\ &2y\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{y^{2}}{x} \\ &2x(\frac{\mathrm{d} y}{\mathrm{d} x})=y \end{aligned}$
Hence,
$\begin{aligned} &2x(\frac{\mathrm{d} y}{\mathrm{d} x})=y \end{aligned}$
is the differential equation corresponding to $y^{2}=4ax$

Differential Equation exercise 21.2 question 3 (ii)

Answer:
$\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}+2\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+x \frac{\mathrm{d} y}{\mathrm{d} x}-y=0$
Hint:
Differentiating the given curve, c- Parameter
Given:
$y=cx+2c^{2}+c^{3}$
Solution:
$y=cx+2c^{2}+c^{3}$
$\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=c+0 \\ &c=\frac{\mathrm{d} y}{\mathrm{d} x} \\ &y=x\frac{\mathrm{d} y}{\mathrm{d} x}+2\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3} \end{aligned}$
$\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}+2\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+x \frac{\mathrm{d} y}{\mathrm{d} x}-y=0$

Differential Equation exercise 21.2 question 3 (iii)

Answer:
$y+x\frac{\mathrm{d} y}{\mathrm{d} x}=0$
Hint:
Differentiating the given equation, where a-Parameter
Given:
$xy=a^{2}$
Solution:
$xy=a^{2}$
$\begin{aligned} &\frac{\mathrm{d} }{\mathrm{d} x}(xy)=\frac{\mathrm{d} }{\mathrm{d} x}(a^{2}) \\ &(1)y+x\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &y+x\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}$

Differential Equation exercise 21.2 question 3 (iv)

Answer:
$\frac{\mathrm{d}^{3} y}{\mathrm{d} x^{3}}=0$
Hint:
Differentiating the given equation
Given:
$y=ax^{2}+bx+c$
Solution:
$y=ax^{2}+bx+c$
$\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(ax^{2}+bx+c) \\ &\frac{\mathrm{d} y}{\mathrm{d} x}=2ax+b \\ &\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=\frac{\mathrm{d} }{\mathrm{d} x}(2ax+b) \end{aligned}$
$\begin{aligned} &\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=2a \\ &\frac{\mathrm{d}^{3} y}{\mathrm{d} x^{3}}=\frac{\mathrm{d} }{\mathrm{d} x}(2a) \\ &\frac{\mathrm{d}^{3} y}{\mathrm{d} x^{3}}=0 \end{aligned}$

Differential Equation exercise 21.2 question 4

Answer:
$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}-4y=0$
Hint:
Differentiating the given equation in first and second order
Given:
$y=Ae^{2x}+Be^{-2x}$
Solution:
$y=Ae^{2x}+Be^{-2x}$
$\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=2Ae^{2x}-2Be^{-2x} \\ &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=4Ae^{2x}+4Be^{-2x} \\ &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=4y \\ &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}-4y=0 \end{aligned}$

Differential Equation exercise 21.2 question 5

Answer:
$\frac{\mathrm{d} ^{2}x}{\mathrm{d} t^{2}}+n^{2}x=0$
Hint:
Differentiating the given equation with respect to x.
Let the same equation and substitutes its value.
Given:
$x=Acos\: nt+Bsin\: nt$
Solution:
$x=Acos\: nt+Bsin\: nt$
$\begin{aligned} &\frac{\mathrm{d} x}{\mathrm{d} t}=-An\: sin(nt)+Bn\: cos(nt) \\ &\frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}= -An^{2}\: cos(nt)+Bn^{2}\: sin(nt) \\ &\frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}=-n^{2}[Acos\: (nt)+Bsin\: (nt)] \end{aligned}$
We get,
$\begin{aligned} &\frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}=-n^{2}x \\ &\frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}+n^{2}x=0 \end{aligned}$

Differential Equation exercise 21.2 question 6

Answer:
$xy\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+x\left ( \frac{\mathrm{d}y }{\mathrm{d} x} \right )^{2}-y\frac{\mathrm{d}y }{\mathrm{d} x}=0$
Hint:
Differentiating the given equation, let it equation (i) and substitute a from (iii) in (ii)
Given:
$y^{2}=a(b-x^{2})$
Solution:
$y^{2}=a(b-x^{2})$
$\begin{aligned} &2y\frac{\mathrm{d} y}{\mathrm{d} x}=a(-2x) \\ &y\frac{\mathrm{d} y}{\mathrm{d} x}=-ax \qquad \dots(ii)\end{aligned}$
Again,
$\begin{aligned} &\frac{\mathrm{d} }{\mathrm{d} x}\left ( y\frac{\mathrm{d} y}{\mathrm{d} x}\right ) =\frac{\mathrm{d} }{\mathrm{d} x}(-ax) \\ &\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=-a \qquad \dots(iii) \end{aligned}$
Substitute equation (iii) in (ii),
$\begin{aligned} &y\frac{\mathrm{d} y}{\mathrm{d} x}=\left [ \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} +y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}\right ]x \\ &y\frac{\mathrm{d} y}{\mathrm{d} x}=x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+xy\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \\ &xy\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-y\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}$

Differential Equation exercise 21.2 question 7

Answer:
$y'^{2}(x^{2}-2y^{2})-4xyy'-x^{2}=0$
Hint:
Differentiating the given equation with respect to x
Given:
$y^{2}-2ay+x^{2}=a^{2}$
Solution:
$y^{2}-2ay+x^{2}=a^{2}$
Differentiating with respect to x
$\begin{aligned} &2y\frac{\mathrm{d} y}{\mathrm{d} x}-2a\frac{\mathrm{d} y}{\mathrm{d} x}+2x=0\\ &y\frac{\mathrm{d} y}{\mathrm{d} x}+x=a\frac{\mathrm{d} y}{\mathrm{d} x} \\ &a=\frac{y\frac{\mathrm{d} y}{\mathrm{d} x}+x}{\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )} \end{aligned}$
Putting the value of a in the given equation, we get
$\begin{aligned} y^{2}-2a\left ( \frac{x+y\frac{\mathrm{d} y}{\mathrm{d} x}}{\frac{\mathrm{d} y}{\mathrm{d} x}} \right )y+x^{2}=\left ( \frac{x+y\frac{\mathrm{d} y}{\mathrm{d} x}}{\frac{\mathrm{d} y}{\mathrm{d} x}} \right )^{2} \end{aligned}$
put
$\begin{aligned} \frac{\mathrm{d} y}{\mathrm{d} x}=y' \end{aligned}$
$\begin{aligned} &y^{2}-2a\left ( \frac{x+yy'}{y'} \right )y+x^{2}=\left ( \frac{x+yy'}{y'} \right )^{2} \\ &\frac{y^{2}y'-2(y^{2}y'+xy)+x^{2}y}{y'}=\frac{y^{2}y'^{2}+2xyy'+x^{2}}{y'^{2}} \end{aligned}$
$\begin{aligned} &y^{2}y'^{2}-2y^{2}y'^{2}-2xyy'+x^{2}y'^{2}-y^{2}y'-2xyy'-x^{2}=0 \\ &-4xyy'+x^{2}y'^{2}-x^{2}-2y^{2}y'^{2}=0 \\ &y'^{2}(x^{2}-2y^{2})-4xyy'-x^{2}=0 \end{aligned}$

Differential Equation exercise 21.2 question 8

Answer:
$\left [ 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]^{3}=r^{2}\left ( \frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right )^{2}$
Hint:
Differentiating the given equation where r is a constant
Given:
$(x-a)^{2}+(y-b)^{2}=r^{2} \qquad \qquad \dots(i)$
Solution:
$(x-a)^{2}+(y-b)^{2}=r^{2}$
$\begin{aligned} &2(x-a)+2(y-b)\frac{\mathrm{d}y }{\mathrm{d} x}=0 \qquad \qquad \dots(ii)\\ &1+(y-b)\frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=0 \\ &(y-b)\frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}}=-\left ( 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ) \qquad \qquad \dots(iii) \end{aligned}$
From equation (ii)
$\begin{aligned} &(x-a)=-(y-b)\frac{\mathrm{d} y}{\mathrm{d} x} \end{aligned}$
Put this value in equation (i)
$\begin{aligned} &\left [ -(y-b)\frac{\mathrm{d} y}{\mathrm{d} x} \right ]^{2}+(y-b)^{2}=r^{2} \qquad \qquad \dots (iv) \end{aligned}$
Squaring equation (iii),
$\begin{aligned} &(y-b)^{2} \left ( \frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right )^{2}=\left ( 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right )^{2} \qquad \qquad \dots(v) \end{aligned}$
Dividing equation (iv) by (v),
$\begin{aligned} &\frac{\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+1}{\left ( \frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right )^{2}}=\frac{r^{2}}{\left [ \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+1 \right ]} \end{aligned}$
$\left [ 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]^{3}=r^{2}\left ( \frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right )^{2}$

Differential Equation exercise 21.2 question 9

Answer:
$(x^{2}-y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}-2xy=0$
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:
$\begin{aligned} &(x-0)^{2}+(y-k)^{2}=k^{2} \\ &x^{2}+(y-k)^{2}=k^{2} \\ &x^{2}+y^{2}-2ky=0 \\ &\frac{x^{2}+y^{2}}{2y}=k \end{aligned}$
$\begin{aligned} &\frac{2y\left ( 2x+2y\frac{\mathrm{d} y}{\mathrm{d} x} \right )-(x^{2}-y^{2})2\frac{\mathrm{d} y}{\mathrm{d} x}}{4y^{2}}=0 \\ &4y\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )-2(x^{2}+y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &4xy=4y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}-2(x^{2}+y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}$
$\begin{aligned} &(4y^{2}-2x^{2}-2y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}+4xy=0 \\ &(2y^{2}-2x^{2})\frac{\mathrm{d} y}{\mathrm{d} x}+2xy=0 \\ &(y^{2}-x^{2})\frac{\mathrm{d} y}{\mathrm{d} x}+2xy=0 \\ &(x^{2}-y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}-2xy=0 \end{aligned}$

Differential Equation exercise 21.2 question 10

Answer:
$x^{2}-y^{2}+2xy\frac{\mathrm{d} y}{\mathrm{d} x}=0$
Hint:
Let the equation of the circle assuming R as radius and (a,b) as the centre of the circle is
$(x-a)^{2}+(y-b)^{2}=R^{2}$
Given:
Circles passes through origin and their centre lie on x-axis. So the centre of the circle will be (a, 0)
$(x-a)^{2}+y^{2}=a^{2} \qquad \qquad \dots (i)$
Solution:
$\begin{aligned} &(x-a)^{2}+y^{2}=a^{2} \\ &2(x-a)dx+2ydy=0 \\ &x-a=-y\frac{\mathrm{d} y}{\mathrm{d} x} \\ &a=x+y\frac{\mathrm{d} y}{\mathrm{d} x} \end{aligned}$
Substituting these values in equation (i)
$\begin{aligned} &x^{2}+y^{2}=2x\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right ) \\ &x^{2}+y^{2}=2x^{2}+2xy\frac{\mathrm{d} y}{\mathrm{d} x} \\ &y^{2}-x^{2}-2xy\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &x^{2}-y^{2}+2xy\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}$

Differential Equation exercise 21.2 question 11

Answer:
$\frac{\mathrm{d} r}{\mathrm{d} t}=-k$
Hint:
Let the surface be A
Rate of evaporation,
$\frac{\mathrm{d} v}{\mathrm{d} t}$
Using formula,
$v=\frac{4}{3}\pi r^{3},\quad A=4\pi r^{2}$
Given:
Rain drop evaporates at a rate proportional to its surface area
Solution:
$\begin{aligned} &\frac{\mathrm{d} v}{\mathrm{d} t}\propto A \\ &\frac{\mathrm{d} v}{\mathrm{d} t}=-kA \\ &\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{4}{3}\pi r^{3} \right )=-k(4\pi r^{2}) \\ &\frac{4}{3}\pi (3r^{2})\frac{\mathrm{d} r}{\mathrm{d} t}=-4\pi r^{2} \\ &\frac{\mathrm{d} r}{\mathrm{d} t}=-k \end{aligned}$

Differential Equation exercise 21.2 question 12

Answer:
$2a\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}=0$
Hint:
Let the equation of parabola be
$(y-k)^{2}=4a(x-h)$
Where h, k are parameter
Given:
Latus rectum 4a and axis are parallel to x-axis
Solution:
Let the equation of parabola be
$(y-k)^{2}=4a(x-h)$
Where h, k are parameter
$\begin{aligned} &2(y-k)\frac{\mathrm{d} y}{\mathrm{d} x}=4a \\ &(y-k)\frac{\mathrm{d} y}{\mathrm{d} x}=2a \\ &(y-k)\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=0 \\ &2a\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}=0 \end{aligned}$
which is the required differential equation.

Differential Equation exercise 21.2 question 13

Answer:
$\frac{\mathrm{d} y}{\mathrm{d} x}+2xy=4x^{3}$
Hint:
$\begin{aligned} &y=2(x^{2}-1)+ce^{-x^{2}} \\ &y=2x^{2}-2+ce^{-x^{2}} \\ &2x^{2}-y=2-ce^{-x^{2}} \\ &\text { Substitutes } 2-ce^{-x^{2}} \text { by } 2x^{2}-y \end{aligned}$
Given:
$\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=4x+ce^{-x^{2}}(-2x) \\ &\frac{\mathrm{d} y}{\mathrm{d} x}=2x(2-ce^{-x^{2}}) \\ &\frac{\mathrm{d} y}{\mathrm{d} x}=2x(2x^{2}-y) \\ &\frac{\mathrm{d} y}{\mathrm{d} x}=4x^{3}-2xy \\ &\frac{\mathrm{d} y}{\mathrm{d} x}+2xy=4x^{3} \end{aligned}$
Hence proved.

Differential Equation exercise 21.2 question 14

Answer:
$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}(1-x^{2})-\frac{\mathrm{d} y}{\mathrm{d} x}x-2=0$
Hint:
Differentiating the given equation with respect to x
Given:
$y=(sin^{-1}x)^{2}+A\, cos^{-1}x+B$
Solution:
$\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=2sin^{-1}x\frac{1}{\sqrt{1-x^{2}}}+A\left ( \frac{-1}{\sqrt{1-x^{2}}} \right )+0\\ &y'=\frac{1}{\sqrt{1-x^{2}}}\left [ 2sin^{-1}x-A \right ]\\ &y'\sqrt{1-x^{2}}=2sin^{-1}x-A\\ &A=2sin^{-1}x-y^{2}\sqrt{1-x^{2}} \end{aligned}$
$\begin{aligned} &0=\frac{2}{\sqrt{1-x^{2}}}-y''\sqrt{1-x^{2}}-y^{2}\frac{1(-2x)}{2\sqrt{1-x^{2}}}\\ &0=\frac{2}{\sqrt{1-x^{2}}}-y''\sqrt{1-x^{2}}+\frac{y^{2}x}{\sqrt{1-x^{2}}}\\ &0=\frac{2-y''(1-x^{2})+y^{3}x}{\sqrt{1-x^{2}}} \end{aligned}$
$\begin{aligned} &y''(1-x^{2})-y'x-2=0 \end{aligned}$
$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}(1-x^{2})-\frac{\mathrm{d} y}{\mathrm{d} x}x-2=0$

Differential Equation exercise 21.2 question 15 (i)

Answer:
$2xy\frac{\mathrm{d} y}{\mathrm{d} x}+4x^{2}-y^{2}=0$
Hint:
Using chain rule and putting the value of a in (i)
Given:
$(2x+a)^{2}+y^{2}=a^{2} \qquad \qquad \dots (i)$
Solution:
$\begin{aligned} &2(2x+a)2+2y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &2(2x+a)+y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &(2x+a)=-\frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x}\\ &a=-\frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x}-2x \end{aligned}$
$\begin{aligned} &\left ( \frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y^{2}=\left ( \frac{y}{2}\frac{\mathrm{d} y}{\mathrm{d} x}+2x \right )^{2}\\ &\left ( \frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y^{2}=\left ( \frac{y}{2}\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+4x^{2}+2\frac{y}{2}.\frac{\mathrm{d} y}{\mathrm{d} x}.2x\\ &y^{2}-4x^{2}-2xy\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}$
$2xy\frac{\mathrm{d} y}{\mathrm{d} x}+4x^{2}-y^{2}=0$

Differential Equation exercise 21.2 question 15 (ii)

Answer:
$2xy\frac{dy}{dx}-(y^{2}+4x^{2})=0$
Hint:
Differentiating the given equation
Given:
$(2x-a)^{2}-y^{2}=a^{2}$
Solution:
$\begin{aligned} &2(2x-a)2-2y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &2x-a=\frac{y}{2}\frac{\mathrm{d} y}{\mathrm{d} x}\\ &a=\left ( 2x-\frac{x}{2}\frac{\mathrm{d} y}{\mathrm{d} x} \right )\\ &\left ( \frac{y}{2}.\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-y^{2}=4x^{2}+\left ( \frac{y}{2}.\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-2.2a.\frac{y}{2}.\frac{\mathrm{d} y}{\mathrm{d} x} \end{aligned}$
$2xy\frac{dy}{dx}-(y^{2}+4x^{2})=0$

Differential Equation exercise 21.2 question 15.3 (iii)

Answer:
$4xy\frac{\mathrm{d} y}{\mathrm{d} x}+x^{2}-2y^{2}=0$
Hint:
Using the formula (a+b)² and differentiating the given equation
Given:
$(x-a)^{2}+2y^{2}=a^{2}$
Solution:
$\begin{aligned} &2(x-a)+2y.2\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &(x-a)=-2y\frac{\mathrm{d} y}{\mathrm{d} x}\\ &a=x+2y\frac{\mathrm{d} y}{\mathrm{d} x}\\ &\left ( 2+\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+2y^{2}=x^{2}+\left ( 2+\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+2x.2y\frac{\mathrm{d} y}{\mathrm{d} x} \end{aligned}$
$4xy\frac{\mathrm{d} y}{\mathrm{d} x}+x^{2}-2y^{2}=0$

Differential Equation exercise 21.2 question 16 (i)

Answer:
The required equation is
$x+y\frac{\mathrm{d} y}{\mathrm{d} x}=0$
Hint:
The given equation is the equation of a circle
Given:
$x^{2}+y^{2}=a^{2}$
Solution:
$x^{2}+y^{2}=a^{2}$
Differentiating with respect to x,
$\begin{aligned} &2x+2y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &x+y\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}$
The required equation is
$x+y\frac{\mathrm{d} y}{\mathrm{d} x}=0$

Differential Equation exercise 21.2 question 16 (ii)

Answer:
The required equation is
$x-y\frac{\mathrm{d} y}{\mathrm{d} x}=0$
Hint:
Differentiating the given equation with respect to x
Given:
$x^{2}-y^{2}=a^{2}$
Solution:
$x^{2}-y^{2}=a^{2}$
Differentiating with respect to x,
$\begin{aligned} &2x-2y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &x-y\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}$
The required equation is
$x-y\frac{\mathrm{d} y}{\mathrm{d} x}=0$

Differential Equation exercise 21.2 question 16 (iii)

Answer:
The required equation is
$y-2x\frac{\mathrm{d} y}{\mathrm{d} x}=0$
Hint:
The equation is the equation of parabola
Given:
$y^{2}=4ax$
Solution:
$y^{2}=4ax$
$\begin{aligned} &2y\frac{\mathrm{d} y}{\mathrm{d} x}=4a\\ &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2a}{y}\\ &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{y}\left ( \frac{y^{2}}{4x} \right )\\ &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{y}{2x}\\ &2x\frac{\mathrm{d} y}{\mathrm{d} x}=y \end{aligned}$
The required equation is
$y-2x\frac{\mathrm{d} y}{\mathrm{d} x}=0$

Differential Equation exercise 21.2 question 16 (iv)

Answer:
The required differential equation is
$x^{2}\left [ 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]=\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$
Hint:
$\text { Putting } \sqrt{1-x^{2}} \text { instead of } (y-b) \text { in }2x+2(y-b)\frac{\mathrm{d} y}{\mathrm{d} x}=0$
Given:
$x^{2}+(y-b)^{2}=1$
Solution:
The equation of family of curves is
$x^{2}+(y-b)^{2}=1 \qquad \qquad \dots(i)$
Where b is a parameter
Differentiating equation (i) with respect to x, we get
$\begin{aligned} &2x+2(y-b)\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &2x+2\sqrt{1-x^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}=0 \qquad \qquad [Using (i)]\\ &x=-\sqrt{1-x^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}\\ &x^{2}=(1-x^{2})\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}\\ &x^{2}=\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-x^{2}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \end{aligned}$
The required equation is
$x^{2}\left [ 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]=\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$

Differential Equation exercise 21.2 question 16 (v)

Answer:
The required differential equation is
$y^{2}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-y^{2}=1$
Hint:
Differentiating equation (i) with respect to x
Given:
$(x-a)^{2}-y^{2}=1$
Solution:
The equation of family of curves is
$(x-a)^{2}-y^{2}=1 \qquad \qquad \dots(i)$
Where a is a parameter
Differentiating equation (i) with respect to x, we get
$\begin{aligned} &2(x-a)-2y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &(x-a)-y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &\sqrt{1+y^{2}}=y\frac{\mathrm{d} y}{\mathrm{d} x} \qquad \qquad [U\! sing (i)]\\ &(1+y^{2})=y^{2}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \end{aligned}$
The required equation is
$y^{2}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-y^{2}=1$

Differential Equation exercise 21.2 question 16 (vi)

Answer:
The required differential equation is
$x\left [ y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]=y\frac{\mathrm{d} y}{\mathrm{d} x}$
Hint:
This is the equation of hyperbola
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Given:
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Solution:
The equation of family of curves is
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \qquad \qquad \dots(i)$
Where a and b are parameter
Differentiating equation (i) with respect to x, we get
$\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}=0 \qquad \qquad \dots(ii)$
Differentiating equation (ii) with respect to x, we get
$\begin{aligned} &\frac{2}{a^{2}}-\frac{2}{b^{2}}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-\frac{2y}{b^{2}}\left ( \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right )=0\\ &\frac{2}{a^{2}}=\frac{2}{b^{2}}\left [ y\left ( \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right )+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]\\ &\frac{b^{2}}{a^{2}}=\left [ y\left ( \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right )+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ] \qquad \qquad \dots (iii) \end{aligned}$
Now, from (ii), we get
$\begin{aligned} &\frac{2x}{a^{2}}=\frac{2y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}\\ &\frac{b^{2}}{a^{2}}=\frac{y}{x}\frac{\mathrm{d} y}{\mathrm{d} x} \qquad \qquad \dots (iv) \end{aligned}$
From (iii) and (iv), we get
$\begin{aligned} &\frac{y}{x}\frac{\mathrm{d} y}{\mathrm{d} x}=\left [ y\left ( \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right )+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ] \end{aligned}$
The required differential equation is
$x\left [ y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]=y\frac{\mathrm{d} y}{\mathrm{d} x}$

Differential Equation exercise 21.2 question 16 (vii)

Answer:
The required differential equation is
$y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=0$
Hint:
Differentiating equation (i) and (ii) with respect to x
Given:
$y^{2}=4a(x-b)$
Solution:
The equation of family of curves is
$y^{2}=4a(x-b) \qquad \qquad \dots (i)$
Where a and b are parameters
Differentiating equation (i) with respect to x, we get
$\begin{aligned} &2y\frac{\mathrm{d} y}{\mathrm{d} x}=4a \\ &y\frac{\mathrm{d} y}{\mathrm{d} x}=2a \qquad \qquad \dots(ii) \end{aligned}$
Differentiating equation (ii) with respect to x, we get
$y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=0$
The required differential equation is
$y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=0$

Differential Equation exercise 21.2 question 16 (viii)

Answer:
The required equation is
$x\frac{\mathrm{d} y}{\mathrm{d} x}=3y$
Hint:
Differentiating the given equation with respect to x
Given:
$y=ax^{3}$
Solution:
$y=ax^{3}$
On differentiating with respect to x, we get
$\frac{\mathrm{d} y}{\mathrm{d} x}=3ax^{2}$
From the given equation,
$a=\frac{y}{x^{3}}$
So, we have
$\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=3\frac{y}{x^{3}}(x^{2})\\ &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3y}{x} \\ &x\frac{\mathrm{d} y}{\mathrm{d} x}=3y \end{aligned}$
The required equation is
$x\frac{\mathrm{d} y}{\mathrm{d} x}=3y$

Differential Equation exercise 21.2 question 16 (ix)

Answer:
The required equation is
$2xy\frac{\mathrm{d} y}{\mathrm{d} x}=(x^{2}+3y^{2})$
Hint:
Differentiating the given equation with respect to x
Given:
$x^{2}+y^{2}=ax^{3}$
Solution:
$\begin{aligned} &x^{2}+y^{2}=ax^{3} \\ &a=\frac{x^{2}+y^{2}}{x^{3}} \end{aligned}$
Differentiating with respect to x,
$\begin{aligned} &\frac{\left [ \left ( 2x+2y\frac{\mathrm{d} y}{\mathrm{d} x} \right )x^{3}-3x^{2}(x^{2}+y^{2}) \right ]}{x^{6}}=0 \\ &2x^{4}+2x^{3}y\frac{\mathrm{d} y}{\mathrm{d} x}-3x^{4}-3x^{2}y^{2}=0 \\ &2x^{3}y\frac{\mathrm{d} y}{\mathrm{d} x}-x^{4}-3x^{2}y^{2}=0 \end{aligned}$
$\begin{aligned} &2x^{3}y\frac{\mathrm{d} y}{\mathrm{d} x}=x^{4}+3x^{2}y^{2} \\ &2x^{3}y\frac{\mathrm{d} y}{\mathrm{d} x}=x^{2}(x^{2}+3y^{2}) \\ &2xy\frac{\mathrm{d} y}{\mathrm{d} x}=(x^{2}+3y^{2}) \end{aligned}$
The required equation is
$2xy\frac{\mathrm{d} y}{\mathrm{d} x}=(x^{2}+3y^{2})$

Differential Equation exercise 21.2 question 16 (x)

Answer:
The required equation is
$x\frac{\mathrm{d} y}{\mathrm{d} x}=y\, log\, y$
Hint:
Differentiating the given equation with respect to x
Given:
$y=e^{ax}$
Solution:
$y=e^{ax}$
Differentiating with respect to x
$\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=ae^{ax} \\ &\frac{\mathrm{d} y}{\mathrm{d} x}=ay \end{aligned}$
From the given equation, we have
$\begin{aligned} &y=e^{ax} \\ &log\, y=ax \\&a=\frac{log\, y}{x} \end{aligned}$
Now,
$\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=ay=y\frac{log\, y}{x} \\ &x\frac{\mathrm{d} y}{\mathrm{d} x}=y\, log\, y\end{aligned}$
The required equation is
$x\frac{\mathrm{d} y}{\mathrm{d} x}=y\, log\, y$

Differential Equation exercise 21.2 question 17

Answer:
The required differential equation is
$xy\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}+x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-y\frac{\mathrm{d} y}{\mathrm{d} x}=0$
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
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \qquad \qquad \dots (i)$
Where a and b are arbitrary constants
Differentiating equation (i) with respect to x, we get
$\begin{aligned} &\frac{2x}{a^{2}}+\frac{2y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &\frac{2x}{a^{2}}=-\frac{2y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x} \\ &\frac{-b^{2}}{a^{2}}=\frac{y}{x}\frac{\mathrm{d} y}{\mathrm{d} x} \qquad \qquad \dots(ii) \end{aligned}$
Differentiating equation (ii) with respect to x, we get
$\begin{aligned} &0=\frac{y}{x}\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\frac{\left ( x\frac{\mathrm{d} y}{\mathrm{d} x}-y \right )}{x^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}\\ &xy\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}+x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-y\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}$
The required differential equation is
$xy\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}+x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-y\frac{\mathrm{d} y}{\mathrm{d} x}=0$

Differential Equation exercise 21.2 question 18

Answer:
The required differential equation is
$x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+xy\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}-y\frac{\mathrm{d} y}{\mathrm{d} x}=0$
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
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \qquad \qquad \dots (i)$
Where a and b are arbitrary constants
Differentiating equation (i) with respect to x, we get
$\begin{aligned} &\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &\frac{x}{a^{2}}-\frac{y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}=0 \qquad \quad \dots(ii) \end{aligned}$
Differentiating equation (ii) with respect to x, we get
$\begin{aligned} &\frac{1}{a^{2}}-\frac{1}{b^{2}}\left [ \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right ]=0 \\ &\frac{1}{a^{2}}=\frac{1}{b^{2}}\left [ \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right ] \end{aligned}$
Substituting this value of 1/a² in equation (ii), we get
$\begin{aligned} &x\left [ \frac{1}{b^{2}}\left [ \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right ] \right ]-\frac{y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &x\left [ \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right ]-y\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+xy\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}-y\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}$
The required differential equation is
$x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+xy\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}-y\frac{\mathrm{d} y}{\mathrm{d} x}=0$

Differential Equation exercise 21.2 question 19

Answer:
The required differential equation is
$\left [ (x+y)^{2}+1 \right ]\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$
Hint:
The equation belongs to the family of a circle
$(x-b)^{2}+(y-k)^{2}=r^{2}$
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
$\begin{aligned} &(x+a)^{2}+(y-a)^{2}=a^{2} \qquad \qquad \dots (i) \\ &x^{2}+y^{2}+2ax-2ay+a^{2}=0\qquad \qquad \dots (ii) \end{aligned}$
Differentiating (ii) with respect to x, we get
$\begin{aligned} &2x+2y\frac{\mathrm{d} y}{\mathrm{d} x}+2a-2a\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &x+y\frac{\mathrm{d} y}{\mathrm{d} x}+a-a\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &x+y\frac{\mathrm{d} y}{\mathrm{d} x}=-a+a\frac{\mathrm{d} y}{\mathrm{d} x} \\ &a=\frac{\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )}{\frac{\mathrm{d} y}{\mathrm{d} x}-1} \end{aligned}$
Substituting this value of a in (i), we get
$\begin{aligned} &\left [ x+\frac{\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )}{\frac{\mathrm{d} y}{\mathrm{d} x}-1} \right ]^{2}+\left [ y-\frac{\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )}{\frac{\mathrm{d} y}{\mathrm{d} x}-1} \right ]^{2}=\left [\frac{\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )}{\frac{\mathrm{d} y}{\mathrm{d} x}-1} \right ]^{2} \end{aligned}$
$\begin{aligned} &\left [ x\left ( \frac{\mathrm{d} y}{\mathrm{d} x}-1 \right ) +\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right ) \right ]^{2}+\left [ y\left ( \frac{\mathrm{d} y}{\mathrm{d} x}-1 \right ) +\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right ) \right ]^{2}=\left [ x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right ]^{2} \end{aligned}$
$\begin{aligned} &(x+y)^{2}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )+(x+y)^{2}=\left [ (x+y)^{2} \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )\right ]^{2} \end{aligned}$
$\left [ (x+y)^{2}+1 \right ]\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$
The required differential equation is
$\left [ (x+y)^{2}+1 \right ]\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$

Differential Equation exercise 21.2 question 20

Answer:
The required differential equation is
$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=\frac{1}{y}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$
Hint:
Differentiating the given equation with respect to x
Given:
$y=ae^{bx+5}$
Solution:
$y=ae^{bx+5} \qquad \qquad \dots(i)$
Where a and b are arbitrary constants
Differentiating with respect to x
$\frac{\mathrm{d} y}{\mathrm{d} x}=a.be^{bx+5} \qquad \qquad \dots (ii)$
Again, Differentiating with respect to x
$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=a.b^{2}e^{bx+5} \qquad \qquad \dots (iii)$
Put
$b=\frac{1}{y}\frac{\mathrm{d} y}{\mathrm{d} x}$
from (i) and (ii) in equation (iii), we get
$\frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}}=y.\frac{1}{y^{2}}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$
$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=\frac{1}{y}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$
The required differential equation is
$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=\frac{1}{y}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$

Differential Equation exercise 21.2 question 21

Answer:
The required differential equation is
$y_{2}-4y_{1}+4y=0$
Hint:
Differentiating the given equation with respect to x
Given:
$y=e^{2x}(a+bx)$
Solution:
$y=e^{2x}(a+bx) \qquad \qquad \dots(i)$
Differentiating with respect to x, we get
$\begin{aligned} &y_{1}=e^{2x}(0+b)+(a+bx)e^{2x}.2 \\ &y_{1}=e^{2x}b+(a+bx)e^{2x}.2 \qquad \qquad \dots(ii) \end{aligned}$
Putting (i) in (ii)
$\begin{aligned} &y_{1}=be^{2x}+2y \\ &y_{1}=2y=be^{2x} \qquad \qquad \dots(iii) \end{aligned}$
Again, Differentiating with respect to x, we get
$\begin{aligned} &y_{2}-2y_{1}=be^{2x}.2 \qquad \qquad \dots(iv) \end{aligned}$
Putting (iii) in (iv), we get
$\begin{aligned} &y_{2}-2y_{1}=2y_{1}-4y \end{aligned}$
The required differential equation is
$y_{2}-4y_{1}+4y=0$

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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