Differential equations - variation of parameters

[accountkiller]

Homework Statement

Find a particular solution using variation of parameters.
y'' + 3y' + 2y = 4e^x

Homework Equations

y_p = -y₁ * INT (y₂f(x)/W[y1,y2]) dx + y₂ * INT (y₁f(x)/W[y1,y2]) dx

The Attempt at a Solution

So, first I find the homogeneous solution, correct?
r² + 3r + 2 = 0, so the roots are - 1 and -2, so
y_h = c₁ * e^-x + c₂ * e^-2x
Then, I use the variation of parameters formula:
y_p1 = -y₁ * INT (y₂f(x)/W[y1,y2]) dx
= 2/3e^x

y_p1 = y₂ * INT (y₁f(x)/W[y1,y2]) dx
= -4e^x

Adding them together, y_p = y_p1 + y_p2 = -10/3e^x.

However, the answer is just 2/3e^x, what I got for y_p1.

 

[tiny-tim]

hi mbradar2!

i think you added instead of subtracting in your determinant for W