Question

The solution of differential equation$x\left(\frac{dy}{dx}\right)=y-x{\cos }^2\left(\frac{y}{x}\right)$ is

• A. $\tan \left(\frac{y}{x}\right)+\log x=c$
• B. $\log \left(\frac{x}{y}\right)+\log x=c$
• C. $\log \left(\frac{x}{y}\right)+\tan x=c$
• D. $\tan \left(\frac{x}{y}\right)+\tan x=c$

Answer 1

The correct option is A

$\tan \left(\frac{y}{x}\right)+\log x=c$

Explanation for the correct option:

Find the solution of the given differential equation

Given:$x\left(\frac{dy}{dx}\right)=y-x{\cos }^2\left(\frac{y}{x}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{y}{x}-{\cos }^2\left(\frac{y}{x}\right)$ ….. $\left(i\right)$

Let $y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$ …… $\left(ii\right)$

Equating, $\left(i\right)$ and $\left(ii\right)$ we get,

$$\begin{gathered} \therefore v+x\frac{dv}{dx}=v-{\cos }^2\left(v\right) \\ \Rightarrow x\frac{dv}{dx}=-{\cos }^2\left(v\right) \\ \Rightarrow \frac{dv}{{\cos }^2\left(v\right)}=-\frac{dx}{x} \\ \Rightarrow {\sec }^2\left(v\right)dv=-\frac{dx}{x} \end{gathered}$$

Integrate both the sides,

$$\begin{gathered} \int {\sec }^2\left(v\right)dv=-\int \frac{dx}{x}\left[\int {\sec }^2\left(x\right)\mathrm{dx}=\tan \left(x\right),\int \frac{1}{x}\mathrm{dx}=\log \left(x\right)\right] \\ \Rightarrow \tan \left(v\right)=-\log \left(x\right)+c \\ \Rightarrow \tan \left(v\right)+\log \left(x\right)=c\left[v=\frac{y}{x}\right] \\ \Rightarrow \tan \left(\frac{y}{x}\right)+\log \left(x\right)=c \end{gathered}$$

Hence, option (A) is the correct answer.