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  • Explain Solution R.D. Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 37 Maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 37 Maths Textbook Solution.

Answers (1)

Answer:\begin{aligned} \sin \left(\frac{y}{x}\right)=\log |x|+\frac{1}{\sqrt{2}} \\ \end{aligned}

Given:\begin{aligned} &x\cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x \end{aligned}

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,

\begin{aligned} &x\cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x \end{aligned}      ....(i)

It is homogeneous equation.

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

So,

\begin{aligned} &x \cos \left(\frac{v x}{x}\right)\left(v+x \frac{d v}{d x}\right)=v x \cos \left(\frac{v x}{x}\right)+x \\ &\Rightarrow x \cos (v)\left(v+x \frac{d v}{d x}\right)=x(v \cos v+1) \\ &\Rightarrow \cos v\left(v+x \frac{d v}{d x}\right)=v \cos v+1 \\ &\Rightarrow v \cos v+x \cos v \frac{d v}{d x}=v \cos v+1 \\ &\Rightarrow x \cos v \frac{d v}{d x}=1 \end{aligned}

Separating the variables and integrating both side we get

\begin{aligned} &\int \cos v d v=\int \frac{d x}{x} \\ &\Rightarrow \sin v=\log |x|+c \\ &\text { Putting } v=\frac{y}{x} \\ &\Rightarrow \sin \left(\frac{y}{x}\right)=\log |x|+c \end{aligned}            ...(ii)

It is given that y=\frac{\pi }{4} when x=1

Putting y=\frac{\pi }{4},  x=1 in equation (ii) we get

\Rightarrow \sin \left ( \frac{\pi }{4} \right )=c

\Rightarrow c=\frac{1}{\sqrt{2}}

Putting value of c in equation (ii) we get

\Rightarrow \sin \left ( \frac{y}{x} \right )=log\mid x\mid +\frac{1}{\sqrt{2}}

This is required solution.

 

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