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  • Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 5 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 5 Maths Textbook Solution.

Answers (1)

Answer: xy=2|y-x|^{\frac{3}{2}}

Given:\frac{dy}{dx}=\frac{y\left ( x+2y \right )}{x\left ( 2x+y \right )},y(1)=2

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

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

Solution: we have,

\frac{dy}{dx}=\frac{y\left ( x+2y \right )}{x\left ( 2x+y \right )},y(1)=2                ...(i)

It is homogeneous equation.

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

So,

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

Separating the variables and integrating both side we get

\begin{aligned} &\int \frac{2+v}{v-v^{2}}=\int \frac{d x}{x}\\ &\int \frac{2+v}{v(v-1)}=\int \frac{d x}{x}\\ &\text { take, }\\ &\frac{2+v}{v(v-1)}=\frac{A}{v}+\frac{B}{v-1}\\ &\Rightarrow \frac{2+v}{v(v-1)}=\frac{A(v-1)+B v}{v(v-1)}\\ &\Rightarrow 2+v=A(v-1)+B v \end{aligned}

comparing the coefficient of like powers of v,

\begin{aligned} &\mathrm{A}=-2 \\ &\mathrm{~A}+\mathrm{B}=1 \Rightarrow \mathrm{B}=3 \\ &\text { Using (ii) :- } \\ &\int\left(\frac{-2}{v}+\frac{3}{v-1}\right) d v=\int \frac{d x}{x} \end{aligned}

\begin{aligned} &\Rightarrow \int \frac{-2}{v} d x+\int \frac{3}{v-1} d v=\int \frac{d x}{x}\\ &\Rightarrow-2 \log |v|+3 \log |v-1|=\log x+\log c\\ &\Rightarrow-\log |\vartheta|^{2}+\log |v-1|^{3}=\log x c\\ &\Rightarrow|v-1|^{3}=v^{2} x c\\ &\text { Putting } \mathrm{v}=\frac{y}{x}\\ &\Rightarrow\left|\frac{y-x}{x}\right|^{3}=\frac{y^{2}}{x} \mathrm{c} \end{aligned}

It is given that y=0 when x=0

Putting y=0, x=1 in equation (iii) we get

\Rightarrow |\frac{2-1}{2}|^{3}=\frac{4}{1}c

\Rightarrow c=\frac{1}{4}

Putting value of c in equation (iii) we get

\Rightarrow \mid \frac{y-x}{x}\mid ^{\frac{3}{2}}=\frac{y^{2}}{x}\left ( \frac{1}{4} \right )

\Rightarrow xy=2\mid y-x\mid ^{\frac{3}{2}}

This is required solution.

 

Posted by

infoexpert21

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