Solve the differential equation `y e^(x/y)dx=(x e^(x/y)+y^2)dy(y!=0)`

Options

A. `e^(x//y)=cosy+C`
B. `e^(x//y)=-siny+C`
C. `e^(y//x)=-cosy+C`
D. `e^(x//y)=-cosy+C`

Answer

C

Explanation

We have,
`ye^(x//y)dx=(xe^(x//y)+y^(2)siny)dy`
`rArr" "(ydx-xdy)e^(x//y)=y^(2)sinydy`
`rArr" "e^(x//y){(ydx-xdy)/(y^(2))}=sin ydy`
`rArr" "e^(x//y)d((x)/(y))=-d(cosy)`
On integrating, we obtain
`e^(x//y)=-cosy+C`, as the required solution.