Y' equals x plus y Explicit solution?

Answer 1

using Laplace transform, we have: sY(s) = Y(s) + 1/(s2) ---> (s-1)Y(s) = 1/(s2), and Y(s) = 1/[(s2)(s-1)]

From the Laplace table, this is ex - x -1, which satisfies the original differential eq.

derivative of [ex - x -1] = ex -1; so, ex - 1 = ex - x - 1 + x

to account for initial conditions, we need to multiply the ex term by a constant C

So y = C*ex - x - 1, and y' = C*ex - 1, with the constant C, to be determined from the initial conditions.