$y'''+3y''+3y'+y = e^{x} -x-1$
Hi, i've got this method of undermined coefficients but i am unable to find the particular solution. I got the complimentary solution as $Y_c = {C_1}{e^{-x}}+ {C_2}{e^{-x}}+ {C_3}{e^{-x}}$ but keep getting the wrong answer when i try to find the particular.
Best Answer
I think you've mistyped the solution to the homogeneous equation. Try:
$$y_h=C_1e^{-x}+C_2xe^{-x}+C_3x^2e^{-x}$$
For the particular solution, try $y=a_1e^x+a_2x+a_3$. Then:
$$ y' = a_1e^x + a_2 $$
$$ y'' = y''' = a_1e^x $$
Then:
$$ \begin{align} y'''+3y''+3y'+y &= 8a_1e^x + a_2x + 3a_2 + a_3 \end{align} $$
and so put
$$ a_1=\frac{1}{8} \qquad a_2=-1 \qquad a_3=2 $$
to get your particular solution as
$$y=\frac{1}{8}e^x-x+2$$