Answer to Question #148807 in Differential Equations for maha

Question

y'=(y/x)+tan(y/x)

Answer 1

Let's introduce the substitution

$$y(x)=x\cdot u(x)\to y'(x)=u(x)+x\cdot u'(x)\longrightarrow\\[0.3cm] u(x)+x\cdot u'(x)=\frac{x\cdot u(x)}{x}+\tan\left(\frac{x\cdot u(x)}{x}\right)\longrightarrow\\[0.3cm] x\cdot u'(x)=u(x)-u(x)+\tan\left(u(x)\right)\longrightarrow\\[0.3cm] \left.x\cdot\frac{du(x)}{dx}=\tan\left(u(x)\right)\right|\times\frac{dx}{x\cdot\tan(u(x))}\longrightarrow\\[0.3cm] \frac{du(x)}{\displaystyle\frac{\sin(u(x))}{\cos(u(x))}}=\frac{dx}{x}\to\frac{\cos(u(x))du(x)}{\sin(u(x))}=\frac{dx}{x}\longrightarrow\\[0.3cm] \int\frac{d\left(\sin(u(x))\right)}{\sin(u(x))}=\int\frac{dx}{x}\to\ln\left|\sin(u(x))\right|=\ln|x|+\ln|C|\\[0.3cm] e^{\ln\left|\sin(u(x))\right|}=e^{\ln|Cx|}\to\sin(u(x))=Cx\to\boxed{u(x)=\arcsin(Cx)}\\[0.3cm] \boxed{y(x)=x\cdot u(x)=x\cdot\arcsin(Cx)}$$

ANSWER

$$y(x)=x\cdot\arcsin(Cx)$$