1.

Solve the differential equation `y e^(x/y)dx=(x e^(x/y)+y^2)dy(y!=0)`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» Correct Answer - C
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.


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