Question

Solve this question y'-7y=sen2x

Answer 1

Multiplying both sides by
`e^(-7x)`, you have


`e^(-7x)y'-7e^(-7x)=e^(-7x)\sin2x`

`(e^(-7x)y)'=e^(-7x)\sin2x`

`e^(-7x)y=\displaystyle\int e^(-7x)\sin2x\,\mathrm dx`

`e^(-7x)y=-\frac1{53} e^(-7 x) (2 cos2x + 7sin2x)+C`

`y=-(2\cos2x+7\sin2x)/(53)+Ce^(7x)`