Question

Find the particular solution of the differential equation $x{e}^{\frac{y}{x}}-ysin\left(\frac{y}{x}\right)+x\frac{dy}{dx}sin\left(\frac{y}{x}\right)=0,$ given that y=0,when x=1.

Answer 1

Given differential equation is homogeneous.
$\therefore y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\text{So,}\frac{dy}{dx}=\frac{ysin\left(\frac{y}{x}\right)-x{e}^{\frac{y}{x}}}{xsin\left(\frac{y}{x}\right)}\Rightarrow v+x\frac{dv}{dx}=\frac{vxsin\left(\frac{vx}{x}\right)-x{e}^{\frac{vx}{x}}}{xsin\left(\frac{vx}{x}\right)}=\frac{vsinv-e^v}{sinv}\Rightarrow v+x\frac{dv}{dx}=v-\frac{e^v}{sinv}orx\frac{dv}{dx}=-\frac{e^v}{sinv}\therefore \int e^{-v}sinvdv=-\int \frac{dx}{x}\Rightarrow I_1=-logx+C_1....\left(i\right)\text{Now}\therefore I_1=\int e^{-v}sinvdv=sinv\int e^{-v}dv+\int e^{-v}cosvdv\Rightarrow =-e^{-v}sinv-e^{-v}cosv-\int e^{v}sinvdv\Rightarrow I_1=-\frac{1}{2}{e}^{-v}(sinv+cosv)+C_2$
Putting value of $I_1$ in (i), $-\frac{1}{2}{e}^{-v}(sinv+cosv)=-log{x}^2+C_1+C_2$
$e^{-\frac{y}{x}}(sin\frac{y}{x}+cos\frac{y}{x})=log{x}^2+C,\text{Where}C=-2C_1-2C_2$
As it is given that y=0,when x=1,so C=1
Hence the solution is $e^{-\frac{y}{x}}(sin\frac{y}{x}+cos\frac{y}{x})=log{x}^2+1$.