Question

Solve the initial value problem.
x y' = y + 7 x^2 sin\(x\), y(4 pi) = 0

Answer 1

`xy'=y+7x^2\sin x`

`xy'-y=7x^2\sin x`

`\frac1xy'-\frac1{x^2}y=7\sin x`

`(\mathrm d)/(\mathrm dx)\left[\frac1xy\right]=7\sin x`

`\frac1xy=\displaystyle\int7\sin x\,\mathrm dx`

`\frac1xy=-7\cos x+C`

`y=-7x\cos x+Cx`

With the initial value
`y(4\pi)=0`, we have


`0=-28\pi+4\piC\implies C=7`

so that the particular solution to the ODE is


`y=-7x\cos x+7x`