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Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (ii) textbook solution.

Answers (1)

Answer :  \sin \frac{y}{x}=\log |x|+c

Hint              : You must know the rules of solving differential equation and integration

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

Solution       :  given differential equation x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x

                       \frac{d y}{d x}=\frac{y \cos \frac{y}{x}+x}{x \cos \frac{y}{x}} \quad \ldots .(i)

It is a homogeneous differential equation

Put y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}

Eqn (i) becomes

\begin{aligned} v+x \frac{d v}{d x} &=\frac{v x \cos v+x}{x \cos v} \\ x \frac{d v}{d x} &=\frac{v \cos v+1}{\cos v}-v \\ x \frac{d v}{d x} &=\frac{v \cos v+1-v \cos v}{\cos v} \end{aligned}

        \begin{aligned} &x \frac{d v}{d x}=\frac{1}{\cos v} \\ &\cos v d v=\frac{d x}{x} \end{aligned}

Integrating both sides

\begin{aligned} &\sin v=\log |x|+c \\ &\sin \frac{y}{x}=\log |x|+c \end{aligned}

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