Question

Solve the differential equation: $\frac{dy}{dx} = \tan (\frac{y}{x})+ \frac{y}{x}$

Answer 1

Please see the attachment

Answer 2

Answer:

$\rm y=x\sin^{-1} (Kx).$

Step-by-step explanation:

$\rm \dfrac{dy}{dx} = \tan \left (\dfrac{y}{x} \right)+ \dfrac{y}{x}.$

This differential equation can be easily solved by putting $\rm t=\dfrac yx$, so that,

$\rm \dfrac{d}{dx}\left ( \dfrac yx \right )  = \dfrac 1x \dfrac{dy}{dx}-\dfrac y{x^2}\\\\\text{On putting the value of t,}\\ \dfrac{dt}{dx} = \dfrac 1x\dfrac {dy}{dx}-\dfrac tx\\x \dfrac{dt}{dx} = \dfrac {dy}{dx}-t\\\dfrac {dy}{dx}=x \dfrac{dt}{dx}+t$

Now, putting the value of t and $\rm \dfrac{dy}{dx}$ in the given differential equation,

$\rm x \dfrac{dt}{dx}+t=\tan t+t\\ x \dfrac{dt}{dx}=\tan t\\\Rightarrow \dfrac{dt}{\tan t}=\dfrac{dx}x\\\\\text{On integrating both sides, }\\\int \dfrac{dt}{\tan t}=\int \dfrac{dx}x\\\int \cot t\ dt=\int \dfrac{dx}x\\\ln (\sin t)=\ln x + C\ \ \ \ \ \ \ \text{where C is the constant of integration.}\\\ln (\sin t)-\ln x = C\\\ln \left (\dfrac{\sin t}{x} \right )=C\\\dfrac{\sin t}{x}=e^C=K\ \ \ \ \ \ \text{where, K is another constant, such that, } e^C = K\\\sin t=Kx$$\rm \sin\left (\dfrac yx \right )= Kx\\\dfrac yx = \sin^{-1} (Kx)\\y=x\sin^{-1} (Kx).$