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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.

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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}


\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


And 2A+B=-1

\Rightarrow B=5


\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.

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