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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.

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

Answers (1)

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

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

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)

Let

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

Posted by

infoexpert21

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