# Use undetermined coefficients to find a particular solution for $y'''+8y'=-8x-3$

Guess is $y = Ax+B$.

$y''' = 0$

$y' = A$

Thus, the differential equation becomes:

$0 + 8(A) = -8x-3$

Where can I go from here? I can't find an explicit solution for A, and my work doesn't even involve the variable B. Any help?

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Well, you started with a guess and got stuck. Which kind of suggests that a different guess might be a way out.... Hint: You're going to need at least some $x$-term on the left-hand side, so maybe a polynomial with degree $\geq 2$ wouldn't be a bad idea. – fgp Oct 20 '12 at 0:49

You'll need $y_p = (Ax+B)x = Ax^2 + Bx$, since the characteristic eqn. of your ODE has $0$ as a root.

$$y'_p = 2Ax + B$$

$$y'''_p = 0$$

So we have:

$$0 + 8(2Ax + B) = -8x - 3$$

And you should be able to take it from there.

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