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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 33 Maths Textbook Solution.

Answer: $x^{2}y^{12}=c^{4}\left ( 2y^{2}x^{2} \right )^{5}$

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

To solve: we have to solve differential equation

Hint: Put  $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Solution: we have,

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

$\Rightarrow \frac{dy}{dx}=\frac{2x^{2}y+y^{3}}{xy^{2}-3x^{3}}$

Put $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$

So,

\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\frac{2 v+v^{3}}{-v^{2}+3}-v \\ &\Rightarrow x \frac{d v}{d x}=\frac{2 v^{3}-v}{-v^{2}+3} \end{aligned}

Separating and Integrating both side we get

\begin{aligned} &\Rightarrow \int \frac{-v^{2}+3}{2 v^{3}-v} d v=\int \frac{1}{x} d x \\ &\therefore \frac{-v^{2}+3}{v\left(2 v^{2}-1\right)}=\frac{A}{v}+\frac{B V+c}{2 v^{2}-1} \text { (using partial fraction) } \\ &\Rightarrow 3-v^{2}=A\left(2 v^{2}-1\right)+(B v+c) v \\ &\Rightarrow 3-v^{2}=(2 A+B) v^{2}+c v-A \end{aligned}

Comparing the coefficient of like power of v, we get

$A=-3,B,C=0$

And $2A+B=-1$

$\Rightarrow B=5$

So,

$\Rightarrow \int \frac{-3}{v} d v+\int \frac{5 v}{2 v^{2}-1} d v=\int \frac{1}{x} d x \\$

$\Rightarrow-3 \int \frac{1}{v} d v+\frac{5}{4} \int \frac{5 v}{2 v^{2}-1} d v=\int \frac{1}{x} d x \\$

$\Rightarrow-3 \log v+\frac{5}{4} \log \left(2 v^{2}-1\right)=\log x+\log c \\$

$\Rightarrow-12 \log v+5 \log \left(2 v^{2}-1\right)=4 \log x+4 \log c \\$

$\Rightarrow \frac{\left(2 v^{2}-1\right)^{2}}{v^{2}}=x^{4} c^{4} \\$

$\Rightarrow \frac{\left[2 y^{2}-x^{2} / x^{2}\right]^{5}}{y^{12} /_{x^{12}}}=x^{4} c^{4}\left[\therefore v=\frac{y}{x}\right] \\$

$\Rightarrow \frac{\left(2 y^{2}-x^{2}\right)^{5} x x^{2}}{y^{2}}=x^{4} c^{4} \\$

$\Rightarrow x^{2} y^{12}=\frac{\left(2 y^{2} x^{2}\right)^{5}}{c^{4}} \\$

$\Rightarrow x^{4} y^{12}=c^{4}\left(2 y^{2} x^{2}\right)^{5}\left[\text { where } \frac{1}{c^{4}}=c^{4}\right]$

This is required solution.