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Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 30 Maths Textbook Solution.

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Answer: \sec \left ( \frac{y}{x} \right )=cxy

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

To solve: We have to solve the given differential equation.

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

Solution: Here, we have,

x \cos \left(\frac{y}{x}\right) \cdot(y d x+x d y)=y \sin \left(\frac{y}{x}\right) \cdot(x d y-y d x)

\Rightarrow \frac{dy}{dx}=\frac{-xy\cos\left ( \frac{y}{x}-y^{2}\sin \left ( \frac{y}{x} \right ) \right )}{-yx\sin \left ( \frac{y}{x}+x^{2}\cos \left ( \frac{y}{x} \right ) \right )}

It is homogeneous equation.

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


\begin{aligned} &\Rightarrow v+x \frac{d v}{d x}=\frac{-x v x \cos \left(\frac{v x}{x}\right)-v^{2} x^{2} \sin \left(\frac{v x}{x}\right)}{-v x^{2} \sin \left(\frac{v x}{x}\right)+x^{2} \cos \left(\frac{v x}{x}\right)} \\ &\Rightarrow x \frac{d v}{d x}=\frac{\left(v \cos v+v^{2} \sin v\right)}{(v \sin v-\cos v)}-v \end{aligned}

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

Separating the variables and integrating both sides we get

\Rightarrow \int \frac{\cos v-v \sin v}{v \cos v} d v=-2 \int \frac{1}{x} d x \\

\Rightarrow \int\left(\frac{1}{v}-\tan v\right) d v=-2 \int \frac{1}{x} d x \\

\Rightarrow \log v-\log |\sec v|=-2 \log x+\log c \\

\Rightarrow \log \left|\frac{v}{|\sec v|}\right|=-\log x^{2}+\log c \\

\Rightarrow \frac{v}{|\sec v|}=\frac{c}{x^{2}} \\

                                                                                                                                        \Rightarrow \frac{\frac{y}{2}}{\sec \left(\frac{y}{x}\right)}=\frac{c}{x^{2}}\left[\therefore v=\frac{y}{x}\right]

\Rightarrow \frac{y}{x \sec \left(\frac{y}{x}\right)}=\frac{c}{x^{2}} \\

\Rightarrow \sec \left(\frac{y}{x}\right)=\frac{x y}{c} \\

\Rightarrow \sec \left(\frac{y}{x}\right)=c x y \text { where } c=\frac{1}{c}

This is required solution.

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