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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 27 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 27 Maths Textbook Solution.

Answers (1)

Answer: \frac{y}{x}+log x =c

Given:\left ( x^{2}-2y^{2} \right )dy+\left ( x^{2}-3xy+2y^{2} \right )dx=0

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

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

Solution: We have,

\left ( x^{2}-2y^{2} \right )dy+\left ( x^{2}-3xy+2y^{2} \right )dx=0

\Rightarrow \frac{dy}{dx}=\frac{x^{2}-3xy+2y^{2}}{2xy-x^{2}}

It is homogeneous equation.

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

So,v+x\frac{dv}{dx}=\frac{x^{2}-3xvx+2x^{2}v^{2}}{2vx^{2}-x^{2}}

\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\frac{8-3 v+2 v^{2}}{1+v^{2}}-v \\ &\Rightarrow x \frac{d v}{d x}=\frac{1-3 v+2 v^{2}}{2 v-1}-v \\ &\Rightarrow x \frac{d v}{d x}=\frac{1-2 v}{2 v-1} \\ &\Rightarrow x \frac{d v}{d x}=-1 \end{aligned}

Separating the variable and integrating both sides, we get,

\begin{aligned} \int d v=-\int \frac{1}{x} d x \\ \end{aligned}

\begin{aligned} &\Rightarrow v=-\log x+c \\ \end{aligned}

\begin{aligned} &\Rightarrow \frac{y}{x}+\log x=c \end{aligned}                                                                                                                                \left [ \therefore v=\frac{y}{x} \right ]

Hence this is required solution.

Posted by

infoexpert21

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