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Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 3 Maths Textbook Solution.

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Answer: log|x|=\cos \frac{y}{x}-1

Given:\frac{dy}{dx}=\frac{y}{x}+\cos ec\frac{y}{x},y\left ( 1 \right )=0

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}{x}+\cos ec\frac{y}{x},y\left ( 1 \right )=0            ....(i)

It is homogeneous equation.

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


\begin{aligned} &v+x \frac{d v}{d x}=\frac{v x}{x}+\operatorname{cosec} \frac{v x}{x} \\ &\Rightarrow x \frac{d v}{d x}=v-\cos e c v-v \\ &\Rightarrow x \frac{d v}{d x}=-\operatorname{cosec} v \\ &\Rightarrow \sin v d v=\frac{d x}{x} \end{aligned}

Separating the variables and integrating both side we get

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

It is given that y=0 when x=0

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

\Rightarrow -\cos \left ( \frac{0}{1} \right )=0+c

\Rightarrow c=-1

Putting value of c in equation (ii) we get

\Rightarrow -\cos \frac{y}{x}=-log|x|+1

\Rightarrow log|x|=\cos \frac{y}{x}-1

This is required solution.


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