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.

## Differential Equations Excercise: 21.2

Differential Equation exercise 21.2 question 1

$8(\frac{\mathrm{dy} }{\mathrm{d} x})^{3}-27y=0$
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

$x\frac{\mathrm{dy} }{\mathrm{d} x}=ylog\: y$
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)

$2x(\frac{\mathrm{d} y}{\mathrm{d} x})=y$
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)

$\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}+2\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+x \frac{\mathrm{d} y}{\mathrm{d} x}-y=0$
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)

$y+x\frac{\mathrm{d} y}{\mathrm{d} x}=0$
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)

$\frac{\mathrm{d}^{3} y}{\mathrm{d} x^{3}}=0$
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

$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}-4y=0$
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

$\frac{\mathrm{d} ^{2}x}{\mathrm{d} t^{2}}+n^{2}x=0$
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

$xy\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+x\left ( \frac{\mathrm{d}y }{\mathrm{d} x} \right )^{2}-y\frac{\mathrm{d}y }{\mathrm{d} x}=0$
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

$y'^{2}(x^{2}-2y^{2})-4xyy'-x^{2}=0$
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

$\left [ 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]^{3}=r^{2}\left ( \frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right )^{2}$
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

$(x^{2}-y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}-2xy=0$
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

$x^{2}-y^{2}+2xy\frac{\mathrm{d} y}{\mathrm{d} x}=0$
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

$\frac{\mathrm{d} r}{\mathrm{d} t}=-k$
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

$2a\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}=0$
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

$\frac{\mathrm{d} y}{\mathrm{d} x}+2xy=4x^{3}$
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

$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}(1-x^{2})-\frac{\mathrm{d} y}{\mathrm{d} x}x-2=0$
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)

$2xy\frac{\mathrm{d} y}{\mathrm{d} x}+4x^{2}-y^{2}=0$
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)

$2xy\frac{dy}{dx}-(y^{2}+4x^{2})=0$
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)

$4xy\frac{\mathrm{d} y}{\mathrm{d} x}+x^{2}-2y^{2}=0$
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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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

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

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

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

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

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.

There is no other option than to practice the whole set of sums multiple times. The top-rated reference book, the RD Sharma Class 12th Exercise 21.2, comes with numerous practice questions for the students to work out. The more sums they practice in differential equations, the better they understand the concept in-depth. With the latest questions updated according to the NCERT syllabus, the Class 12 RD Sharma Chapter 21 Exercise 21.2 Solution is the best reference guide. For the sums that have various possible methods given in this book, students can select the ones that they feel to adapt.

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## RD Sharma Chapter-wise Solutions

1. Which reference books are very much helpful for the students who struggle with the chapter Differential Equation?

The RD Sharma Class 12th Exercise 21.2 reference book is the best guide for the students who are unclear about the sums in the Differential Equation chapter.

2. Where can the students find the RD Sharma books that can be accessed without payment?

Everyone can access the RD Sharma books at the Career 360 website without any payment.

3. How can a student improve their ability to solve the Differential Equation sums without any mistakes?

Practice is the best way that a student can follow to understand the sums in the Differential Equation chapter. They can use the RD Sharma Class 12th Exercise 21.2 solution material to guide them through this process.

4. What is the most common reference guide for class 12 mathematics used by the CBSE board school students?

The CBSE board school students' most preferred and used mathematics guide is that of the RD Sharma collections.

5. How many questions should a student solve for chapter 21 in class 12 mathematics?

There are 27 questions of chapter 21; mathematics asked in the textbook; along with these questions, the students must work out more additional sums to gain practice.

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