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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 33 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 33 Maths Textbook Solution.

Answers (1)

Answer: x^{2}y^{12}=c^{4}\left ( 2y^{2}x^{2} \right )^{5}

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

To solve: we have to solve differential equation

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

Solution: we have,

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

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

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

So,

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

Separating and Integrating both side we get

\begin{aligned} &\Rightarrow \int \frac{-v^{2}+3}{2 v^{3}-v} d v=\int \frac{1}{x} d x \\ &\therefore \frac{-v^{2}+3}{v\left(2 v^{2}-1\right)}=\frac{A}{v}+\frac{B V+c}{2 v^{2}-1} \text { (using partial fraction) } \\ &\Rightarrow 3-v^{2}=A\left(2 v^{2}-1\right)+(B v+c) v \\ &\Rightarrow 3-v^{2}=(2 A+B) v^{2}+c v-A \end{aligned}

Comparing the coefficient of like power of v, we get

A=-3,B,C=0

And 2A+B=-1

\Rightarrow B=5

So,

\Rightarrow \int \frac{-3}{v} d v+\int \frac{5 v}{2 v^{2}-1} d v=\int \frac{1}{x} d x \\

\Rightarrow-3 \int \frac{1}{v} d v+\frac{5}{4} \int \frac{5 v}{2 v^{2}-1} d v=\int \frac{1}{x} d x \\

\Rightarrow-3 \log v+\frac{5}{4} \log \left(2 v^{2}-1\right)=\log x+\log c \\

\Rightarrow-12 \log v+5 \log \left(2 v^{2}-1\right)=4 \log x+4 \log c \\

\Rightarrow \frac{\left(2 v^{2}-1\right)^{2}}{v^{2}}=x^{4} c^{4} \\

\Rightarrow \frac{\left[2 y^{2}-x^{2} / x^{2}\right]^{5}}{y^{12} /_{x^{12}}}=x^{4} c^{4}\left[\therefore v=\frac{y}{x}\right] \\

                                                                                        \Rightarrow \frac{\left(2 y^{2}-x^{2}\right)^{5} x x^{2}}{y^{2}}=x^{4} c^{4} \\

                                                                                       \Rightarrow x^{2} y^{12}=\frac{\left(2 y^{2} x^{2}\right)^{5}}{c^{4}} \\

\Rightarrow x^{4} y^{12}=c^{4}\left(2 y^{2} x^{2}\right)^{5}\left[\text { where } \frac{1}{c^{4}}=c^{4}\right]

This is required solution.

Posted by

infoexpert21

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