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  • Please Solve R.D.Sharma class 12 Chapter 21 Differential Equations Exercise 21.9 Question 11 Maths textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 11 Maths textbook Solution.

Answers (1)

Answer:  \tan ^{-1}\left ( \frac{y}{x} \right )=c+log|x|

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

To find: we have to find the solution of 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}+xy+y^{2}

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

It is a homogeneous equation of degree 2.

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

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

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

\begin{aligned} &\tan ^{-1} v=\log x+c \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \int \frac{d v}{1+v^{2}}=\tan ^{-1} v\right] \\ &\Rightarrow \tan ^{-1} \frac{y}{x}=\log x+c \end{aligned}                                                        \left [ \therefore v=y/x \right ]

This is required solution.

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