I've got a problem with finding particular solution of: $$y''-y'-6y=12x.$$ My homogeneous solution is : $$y=C_1e^{3x}+C_2e^{-2x}.$$

When i'm trying to find particular solution i'm using the Method of undetermined coefficients.

My solution here is : $$y_{\text{particular}} = -2x$$

And the answer i should get is: $$y_{\text{particular}} = -2x+\frac{1}{3}.$$

How can i get that?

Thanks in advance.

  • 1
    $\begingroup$ try a particular solution in the form $y+p = a + bx.$ $\endgroup$ – abel May 17 '15 at 15:02

Searching a particular solution of the form $y=ax+b$ we find:

$$ y'=a\quad y''=0 \Rightarrow -a-6(ax+b)=12x \iff(-6b-a)-6ax=12x $$ so: $-6b-a =0$ and $-6a=12$ and $a=-2$, $b=1/3$.


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