#### Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 1 Maths textbook Solution.

Answer: $x^{2}y=\left ( y+2x \right )$

Given:  Here given that $x^{2}dy+y\left ( x+y \right )dx=0$

To solve: We have to solve the given differential equation.

Hint: In homogeneous differential equation put $y=vx$ and

Solution: We have,

$x^{2}dy+y\left ( x+y \right )dx=0$

$\Rightarrow \frac{dy}{dx}=\frac{y\left (x+y \right )}{x^{2}}$

Clearly since each of the functions $xy+y^{2}$is and $x^{2}$ homogeneous function of degree $c_{i}$ the given equation is homogeneous.

Putting $y=vx$ and $\frac{dy}{dx}=v+\frac{xdv}{dx}$the given

Equation becomes

\begin{aligned} &v+\frac{x d v}{d x}=-\frac{v x(x+v x)}{x^{2}} \\ &\Rightarrow v+x \frac{d v}{d x}=-\frac{v x^{2}(1+v )}{x^{2}} \\ &\Rightarrow v+\frac{x d v}{d x}=-v-v^{2} \\ &\Rightarrow x \frac{d v}{d x}=-\left(2 v+v^{2}\right) \\ &\Rightarrow \frac{d v}{v^{2}+2 v} d v=-\frac{d x}{x} \end{aligned}

Integrating on both side, we get

\begin{aligned} &\int \frac{dv}{v^{2}+2 v}=-\int \frac{d x}{x} \\ &\Rightarrow \int \frac{dv}{v^{2}-2 v+1-1}=-\int \frac{d x}{x} \\ &\Rightarrow \int \frac{dv}{(v-1)^{2}-(1)^{2}}=\int \frac{d x}{x} \\ &\Rightarrow \frac{1}{2} \log \left|\frac{v+1-1}{v+1+1}\right|=-\log x+\log c \\ &\Rightarrow \log \left|\frac{v}{v+2}\right|^{1 / 2}=-\log \left|\frac{c}{x}\right|\left[\therefore \log a-\log b=\left.\log \right|^{a} / b \mid\right] \\ &\Rightarrow\left(\frac{v}{v+2}\right)^{1 / 2}=\frac{c^{2}}{x^{2}} \end{aligned}

\begin{aligned} &\therefore y=v x \Rightarrow v=\frac{y}{x} \\ &\Rightarrow \frac{y / x}{\frac{y}{x}+2}=\frac{c^{2}}{x^{2}} \\ &\Rightarrow \frac{y}{y+2 x}=\frac{c^{2}}{x^{2}} \\ &\Rightarrow x^{2} y=c^{2}(y+2 x) \\ &\Rightarrow x^{2} y=c(y+2 x) \end{aligned}                                                                                    where C=c2

This is required solution.