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  • Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 7 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 7 Maths Textbook Solution.

Answers (1)

Answer: x^{4}+6x^{2}y^{2}+y^{4}=8

Given: x\left(x^{2}+3 y^{2}\right) d x+y\left(y^{2}+3 x^{2}\right) d y=0, y(1)=!

To find: we have to find the solution of given differential equation.

Hint: we will put y=vx and \frac{dy}{dx}=v+x\frac{dv}{dx}

Solution: we have,

x\left(x^{2}+3 y^{2}\right) d x+y\left(y^{2}+3 x^{2}\right) d y=0, y(1)=!

\frac{dy}{dx}=\frac{x\left ( x^{2}+3y^{2} \right )}{y\left ( y^{2}+3x^{2} \right )}

It is homogeneous equation.

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

So,

\begin{aligned} &v+x \frac{d v}{d x}=\frac{x\left(x^{2}+3 y^{2}\right)}{y\left(y^{2}+3 x^{2}\right)} \\ &\Rightarrow x \frac{d v}{d x}=-\frac{x\left(x^{2}+3 v^{2} x^{2}\right)}{v x\left(v^{2} x^{2}+3 x^{2}\right)}-v \\ &\Rightarrow x \frac{d v}{d x}=-\frac{1+3 v^{2}}{v\left(v^{2}+3\right)}-v \\ &\Rightarrow x \frac{d v}{d x}=\frac{-v^{4}-6 v^{2}-1}{v\left(v^{2}+3\right)} \end{aligned}

Separating the variables and integrating both side we get

\begin{aligned} &\int \frac{v\left(v^{2}+3\right)}{v^{4}+6 v^{2}+1}=-\int \frac{d x}{x} \\ &\Rightarrow \int \frac{4 v\left(v^{2}+3\right)}{v^{4}+6 v^{2}+1}=-4 \int \frac{d x}{x} \quad[\text { Multiply4] } \\ &\Rightarrow \log \left|v^{4}+6 v^{2}+1\right|=-4 \log |x|+\log |c| \\ &\Rightarrow \log \left|v^{4}+6 v^{2}+1\right|=\log \left|\frac{c}{x^{4}}\right| \\ &\Rightarrow\left|v^{4}+6 v^{2}+1\right|=\left|\frac{c}{x^{4}}\right| \end{aligned}

Putting v=\frac{y}{x}

\begin{aligned} &\Rightarrow\left|\frac{y^{4}}{x^{4}}+6 \frac{y^{2}}{x^{2}}+1\right|=\left|\frac{c}{x^{4}}\right|\\ &\Rightarrow\left|y^{4}+6 x^{2} y^{2}+1\right|=|c| \end{aligned}        ....(iii)

It is given that y(1)=!

Putting y=1, x=1 in equation (ii) we get

\Rightarrow 1+6+1=c

\Rightarrow c=8

Putting value of c in equation (ii) we get

    \begin{aligned} &\Rightarrow\left|y^{4}+6 x^{2} y^{2}+1\right|=8 \\ &\Rightarrow y^{4}+6 x^{2} y^{2}+1=8 \end{aligned}

This is required solution.

Posted by

infoexpert21

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