Question

Y^-1 * dy+y e^cosx * sinx * dx=0

Answer 1

`\frac{\mathrm dy}y+ye^(\cos x)\sin x\,\mathrm dx=0`

`\implies(\mathrm dy)/(y^2)=-e^(\cos x)\sin x\,\mathrm dx`


`\displaystyle\int(\mathrm dy)/(y^2)=-\int e^(\cos x)\sin x\,\mathrm dx`

`-\frac1y=e^(\cos x)+C`

`y=-\frac1{e^(\cos x)+C}`