Answer to Question #222718 in Differential Equations for liam

Question

y''-3y'+2y=2x²+e^x+2xe^x+4e^3x

Answer 1

$\displaystyle \textsf{The auxiliary equation is}\\ m^2 - 3m + 2 = 0\\ m = 1, 2\\ \implies y = C_1e^x + C_2e^{2x}\\ y''-3y'+2y=2x^2+e^x+2xe^x+4e^(3x)\\ y = Ax^2 + Bx + C + Dxe^x + Ex^2e^x + Fe^{3x}\\ y' = 2Ax + B + De^x + Dxe^x + 2Exe^x + Ex^2e^x + 3Fe^{3x}\\ \begin{aligned} y'' &= 2A + 2De^x + Dxe^x + 2Ee^x + 2Exe^x \\&+ Ex^2e^x + 2Exe^x + 9Fe^{3x} \\&= 2A + (2D + 2E)e^x + (D + 4E)xe^x + Ex^2e^x + 9Fe^{3x} \end{aligned}\\ y' = 2Ax + B + De^x + (D + 2E)xe^x + Ex^2e^x + 3Fe^{3x}\\ y = Ax^2 + Bx + C + Dxe^x + Ex^2e^x + Fe^{3x}\\ \begin{aligned} y" - 3y' + 2y &= (2A - 3B + 2C) + (2E - D)e^x - 2Exe^x \\&+ 2Fe^{3x} + (2B - 6A)x + 2Ax^2 \\&= 2x^2+e^x+2xe^x+4e^(3x) \end{aligned}\\ 2A = 2, A = 1\\ 2B - 6A = 0, B = 3\\ 2F = 4, F = 2\\ -2E = 2, E = -1\\ 2E - D = 1, D = -3\\ 2A - 3B + 2C = 0\\ 2 - 9 + 2C = 0, C = 7/2\\ y = x^2 + 3x + 7/2 - 3xe^x - x^2e^x + 2e^{3x}\\ \implies y = C_1e^x + C_2e^{2x} + x^2 + 3x \\+ 7/2 - 3xe^x - x^2e^x + 2e^{3x}$