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Provide Solution For R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 13 Maths Textbook Solution.

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Answer: 3x^{2}y+2y^{3}=c,xy\neq 0

Given:     2xydx+\left ( x^{2}+2y^{2} \right )dy=0

To solve: we have to solve given differential equation

Hint: In homogeneous differential equation put  y=vx and \frac{dy}{dx}=v+x\frac{dv}{dx}

or put x=yv\Rightarrow \frac{dx}{dy}=v+x\frac{dv}{dy}

Solution: we have,

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


This is homogeneous equation

\text { Put } x=y v \Rightarrow \frac{d x}{d y}=v+x \frac{d v}{d y}\\

\text { (1) } \Rightarrow v+x \frac{d v}{d y}=\frac{v^{2} y^{2}+2 y^{2}}{-2 v y^{2}}\\

=\frac{y^{2}\left(v^{2}+2\right)}{-2 v y^{2}}\\

\Rightarrow y \frac{d x}{d y}=\frac{v^{2}+2}{-2 v}-v\\

\Rightarrow y \frac{d x}{d y}=\frac{3 v^{2}+2}{-2 v}

Separating the variables we get

\int \frac{2v}{-3v^{2}+2}dv=-\int \frac{dy}{y}=         (Integrating \; both \; side)


I=\int \frac{2 v}{-3 v^{2}+2} \\

\text { Put } t=3 v^{2}+2 \\

\Rightarrow d t=6 v d v \\

\Rightarrow \frac{d t}{3}=2 v d v \\

\Rightarrow I=\frac{1}{3} \int \frac{d t}{t}=\frac{1}{3} \log t=\log t^{1 / 3}

\Rightarrow t^{1 / 3}=\frac{c}{y} \\

\Rightarrow \log t^{1 / 3}=\log y+\log c \\

\Rightarrow t=\frac{c^{3}}{y^{3}} \\

\Rightarrow\left(3 v^{2}+2\right)=\frac{c^{3}}{y^{3}}\left[\therefore t=3 v^{2}+2\right] \\

\Rightarrow y^{3}\left(\frac{3 x^{2}}{y^{2}}+2\right)=c^{3}\left[\therefore v=\frac{x}{y}\right]

\begin{aligned} &\Rightarrow y^{3}\left(\frac{3 x^{2}+2 y^{2}}{y^{2}}\right)=c^{3} \\ &\Rightarrow\left(3 x^{2} y+2 y^{3}\right)=c \quad \text { where } C=c^{3} \end{aligned}

This is required solution

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