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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.

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

Answers (1)

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}

So,

\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.

Posted by

infoexpert21

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