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  • Please Solve R.D. Sharma class 12 Chapter 21 Differential Equations Exercise 21.9 Question 8 Maths textbook Solution.

Please Solve R.D. Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 8 Maths textbook Solution.

Answers (1)

Answer: \frac{x+\sqrt{2y}}{x-\sqrt{2y}}=\left ( cx^{2} \right )^{\sqrt{2}}

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

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

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

Solution: we have,

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

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

This is a homogeneous differential equation.

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

We have,

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

\Rightarrow v+x \frac{d v}{d x}=1-2 v^{2}+v \\

\Rightarrow \frac{d v}{1-2 v^{2}}=\frac{d x}{x} \\

\Rightarrow \frac{d v}{v^{2}-\frac{1}{2}}=-2 \frac{d x}{x} \\

\Rightarrow \int \frac{d v}{\left(\frac{1}{\sqrt{2}}\right)^{2}-v^{2}}=2 \int \frac{d x}{x} \\

\Rightarrow \frac{\sqrt{2}}{2} \log \left|\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}\right|=2 \log x+\log c \\

\left.\Rightarrow \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\right] \\

\Rightarrow \frac{1}{\sqrt{2}} \log \left(\frac{\frac{1}{\sqrt{2}}+{ }^{y} / x}{\frac{1}{\sqrt{2}}-{ }^{y} /{x}}\right)=2 \log x+\log c

\begin{aligned} &\Rightarrow \frac{1}{\sqrt{2}} \log \left(\frac{x+y \sqrt{2}}{x-y \sqrt{2}}\right)=\log x^{2}+\log c \\ &\Rightarrow \log \left(\frac{x+y \sqrt{2}}{x-y \sqrt{2}}\right)^{\frac{1}{\sqrt{2}}}=\log c x^{2} \\ &\Rightarrow \frac{x+y \sqrt{2}}{x-y \sqrt{2}}=\left(c x^{2}\right)^{\sqrt{2}} \end{aligned}

Hence this is required solution.

 

Posted by

infoexpert21

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