Find a particular solution of the differential equation $-3y''-2y'+y=3xe^x$

Using the method of undetermined coefficients.

Guess $(Ax+B)e^x$

Plug into diff eq:

$-3[(Ax+B)e^x]'' - 2[(Ax+B)e^x]' + (Ax+B)e^x = 3xe^x$

Wolfram alpha simplifies this to: $A(x-2)=e^x(4B+3x)$. Solving for A and B we find $A = 3e^3$ and $B = -6/4$.

Thus, we have $y_p = [3xe^3-(6/4)]e^x$ but this is incorrect. What am I doing wrong here?

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You must solve the equation for all $x$, leave $x$ a unknown. There is no such solution for the simplified equation you got from wolfram alpha, since the $e^x$ term gets in the way. Are you sure you entered the equatin correctly? –  fgp Oct 20 '12 at 0:09

I get

$$A=-\frac{3}{4}\;\;,\;\;B=\frac{3}{2}$$

and everything's peachy...So either you fed WA with wrong data or WA is wrong.

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$$(Axe^x+Be^x)'=Axe^x+(A+B)e^x$$

$$(Axe^x+Be^x)''=Axe^x+(2A+B)e^x$$

Using that, I get

$$-3A-2A+A=-4A=3$$

$$-3(2A+B)-2(A+B)+B=-8A-4B=0$$

$$B=-2A$$

That seems to agree with DonAntonio's solution.

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