Here's an alternative approach to working the problem...we are given:
$$y''+6y'=-4xe^{-6x}$$
Let:
$$u=y'\implies u'=y''$$
And we have:
$$u'+6u=-4xe^{-6x}$$
Now, if we multiply through by an integrating factor of:
$$\mu(x)=e^{6x}$$
We obtain:
$$e^{6x}u'+6e^{6x}u=-4x$$
The LHS may now be rewritten as the differentiation of a product:
$$\frac{d}{dx}\left(e^{6x}u\right)=-4x$$
Integrate w.r.t $x$:
$$e^{6x}u=-2x^2+c_1$$
And so we have:
$$u=-2x^2e^{-6x}+c_1e^{-6x}$$
Back-substitute for $u$:
$$y'=-2x^2e^{-6x}+c_1e^{-6x}$$
Integrate w.r.t $x$:
$$y(x)=\int -2x^2e^{-6x}+c_1e^{-6x}\,dx=-2\int x^2e^{-6x}\,dx-\frac{1}{6}c_1e^{-6x}$$
Now, if we note that the factor in the second term of $$-\frac{1}{6}c_1$$ is just an arbitrary constant, we may write:
$$y(x)=-2\int x^2e^{-6x}\,dx+c_2e^{-6x}$$
Now, on the remaining integral, let's use IBP, where:
$$u=x^2\implies du=2x\,dx$$
$$dv=e^{-6x}\,dx\implies v=-\frac{1}{6}e^{-6x}$$
And so we have:
$$y(x)=-2\left(-\frac{1}{6}x^2e^{-6x}+\frac{1}{3}\int xe^{-6x}\,dx\right)+c_2e^{-6x}=\frac{1}{3}x^2e^{-6x}-\frac{2}{3}\int xe^{-6x}\,dx+c_2e^{-6x}$$
For the remaining integral, let's use IBP again, where:
$$u=x\implies du=dx$$
$$dv=e^{-6x}\,dx\implies v=-\frac{1}{6}e^{-6x}$$
And so we have:
$$y(x)=\frac{1}{3}x^2e^{-6x}-\frac{2}{3}\left(-\frac{1}{6}xe^{-6x}+\frac{1}{6}\int e^{-6x}\,dx\right)+c_2e^{-6x}=\frac{1}{3}x^2e^{-6x}+\frac{1}{9}xe^{-6x}+\frac{1}{18^2}e^{-6x}+c_1+c_2e^{-6x}$$
$$y(x)=c_1+\left(c_2+\frac{1}{18^2}\right)e^{-6x}+\frac{x}{9}\left(3x+1\right)e^{-6x}$$
Noting that $$\left(c_2+\frac{1}{18^2}\right)$$ is still just an arbitrary constant, we may write:
$$y(x)=c_1+c_2e^{-6x}+\frac{x}{9}\left(3x+1\right)e^{-6x}$$
And this is equivalent to the solution obtain through the method of undetermined coefficients (but with a lot more work). :D
)...you should have:
