Answer to Question #276148 in Differential Equations for pshhh

Question

(D^3 + 3D^2-D-3)y = e^2x+4

Answer 1

Solution

For general solution of the homogeneous equation y’’’+ 3y’’- y’- 3y = 0 we need to solve characteristic equation

k³ + 3k² – k – 3 = 0

(k + 3)(k² – 1) = 0

(k + 3)(k – 1) (k + 1) = 0  => k₁ = -3, k₂ = 1, k₃ = -1

So general solution of the homogeneous equation is y₀(x) = Ae^-3x + Be^x + Ce^-x , where A, B, C are arbitrary constants.

Let’s find partial solution of nonhomogeneous equation in the form y₁(x) = Ee^2x + F, where E, F are arbitrary constants. Substitution into equation gives

y₁’ = 2Ee^2x , y₁’’ = 4Ee^2x , y₁’’’ = 8Ee^2x

y₁’’’+ 3y₁’’- y₁’- 3y₁ = 8Ee^2x + 12Ee^2x – 2Ee^2x – 3Ee^2x – 3F = 15Ee^2x  – 3F = e^2x+4

So E = 1/15, F = -4/3

Therefore solution of the given equation is

y(x) = y₀(x) + y₁(x) = Ae^-3x + Be^x + Ce^-x + e^2x /15 – 4/3