Solve y’’’ + y'= sec x.

Answer 1

y' = u; u' = y'' and u'' = y''', and we get second order equation: u'' + u = secx (1);.

Characteristic equation: r² + 1 = 0; r = ±i; u₁ = cosx; u₂ = sinx. General solution of Homogenius ODE:

u_g = C₁cosx + C₂sinx.

Wronskian W = det{(cosx sinx) (- sinx cosx)] = cos²x + sin²x = 1.

By variation of parameters method particular solution of ODE (1):

u_p = - u₁∫u₂secx/Wdx + u₂∫u₁secx/Wdx = - cosx∫sinx/cosxdx + sinx∫dx = cosxlnIcosxI + xsinx.

So, u = y' = C₁cosx + C₂sinx + cosxlnIcosxI + xsinx; y = ∫(C₁cosx + C₂sinx + cosxlnIcosxI + xsinx)dx =

C₁sinx - C₂cosx + sinxlnIcosxI + lnItanx + secxI - xcosx + C₃