Find sol'n of nonhomogenous differential equation

[accountkiller]

Homework Statement

A nonhomogeneous differential equation, a complimentary solution y_c, and a particular solution y_p are given. Find a solution satisfying the initial conditions.

y'' + y = 3x, y(0) = 2, y'(0) = -2, y_c = c₁cos(x) + c₂sin(x), y_p = 3x.

Homework Equations

y = y_p + c₁y₁ + ... + c_ny_n

The Attempt at a Solution

So I first tried solving for the associated homogenous equation y'' + y = 0.
Guessing that y = e^rx, y' = re^rx and y'' = r²e^rx, so
y'' + y = r²e^rx + e^rx = e^rx(r²+1) = 0, so
r² + 1 = 0...
but that gives me a square root of a negative number for r?

For complex numbers, all I have in my notes from lecture is this example:
y₁ = e^r₁x and y₂ = e^r₂x with r₁ = A + Bi. Solve the DE.
y'' - 2Ay' + (A² + B²)y = 0.

I'm not sure how to use that.

 

[accountkiller]

Wait, since y = y_p + c₁y₁ + c₂y₂ and I know what y_p and y₁ and y₂ are, I can just plug it in, use the initial conditions to solve for c₁ and c₂, and I get the correct answer, which is y = 2cos(x) - 5sin(x) + 3x.

I don't have to do anything with the roots?
In a similar example problem the teacher did, he used roots (although not complex roots like this one has).

 

[Mark44]

Wait, since y = y_p + c₁y₁ + c₂y₂ and I know what y_p and y₁ and y₂ are, I can just plug it in, use the initial conditions to solve for c₁ and c₂, and I get the correct answer, which is y = 2cos(x) - 5sin(x) + 3x.

Right - they have already told you what the solutions are, so there's no point in doing the work to find them.

I don't have to do anything with the roots?
In a similar example problem the teacher did, he used roots (although not complex roots like this one has).