It is given that y = e^x(acosx + bsinx) ------(1)

Now, differentiating both w.r.t. x, we get,

y’ = e^x(acosx + bsinx) + e^x(-asinx + bcosx)

⇒ y’ = e^x[(a + b)cosx – (a – b)sinx)] ------(2)

Again, differentiating both sides w.r.t. x, we get,

y” = e^x[(a + b)cosx – (a – b)sinx)] + e^x[-(a + b)sinx – (a – b)cosx)]

⇒ y” = e^x[2bcosx – 2asinx]

⇒ y” = 2e^x(bcosx – asinx) ----(3)

Adding equation (1) and (3), we get,

⇒ 2y + y” = 2y’

Therefore, the required differential equation is 2y + y” = 2y’= 0.