The method of undetermined coefficients (II)

Problem:

y'' + y' = x

Solution:

Again, we solve the homogeneous equation $z'' + z' = 0$, the solution is $z = Ae^{-x} + B$.

Next we guess the particular solution to $w'' + w' = x$ is a quadratic polynomial $w(x) = Px^2 + Qx + R$, the derivatives are

$w'(x) = 2Px + Q$
$w''(x) = 2P$.

Therefore we have $2P + 2Px + Q = x$, so we get $P = \frac{1}{2}$ and $Q = -1$.

The final answer is therefore $y = Ae^{-x} + \frac{1}{2}x^2 - x + B$.