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  • Need Solution for R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 31 Maths Textbook Solution.

Need Solution for R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 31 Maths Textbook Solution.

Answers (1)

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

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

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

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

Solution: we have,

\begin{aligned} &\left(x^{2}+3 x y+y^{2}\right) d x-x^{2} d y=0 \\ &\Rightarrow \frac{d y}{d x}=\frac{x^{2}+3 x y+y^{2}}{x^{2}} \end{aligned}

It is a homogeneous equation Put    y=vx  and \frac{dy}{dx}=v+x\frac{dv}{dx}

So,

\begin{aligned} &x \frac{d v}{d x}+v=\frac{x^{2}+3 x^{2}+v^{2} x^{2}}{x^{2}} \\ &\Rightarrow x \frac{d v}{d x}+v=\frac{1+3 v+v^{2}}{1}-v \\ &\Rightarrow x \frac{d v}{d x}+v=1+2 v+v^{2} \\ &\Rightarrow x \frac{d v}{d x}+v=(1+v)^{2}\left[\therefore(v+1)^{2}=1+2 v+v^{2}\right] \end{aligned}

 

Separating the variable and Integrating bot side we get

\begin{aligned} &\Rightarrow \int \frac{1}{(v+1)^{2}} d v=\int \frac{d x}{x} \\ &-\frac{1}{v+1}= \log x+k\left[\therefore \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &\Rightarrow \frac{-1}{y / x+1}=\log x+k \\ &\Rightarrow \frac{-x}{x+y}=\log x+k \\ &\Rightarrow \frac{x}{x+y}=\log x=-k \\ &\Rightarrow \frac{x}{x+y}=\log x=c \quad \quad \text { where } c=-\mathrm{k} \end{aligned}

This is a required solution

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