Im trying to find the particular solution to: $$y'' – 3y' + 2y = 2e^x$$
I already have the homogenous solution so this is not my problem.
Assuming that $y_p = Ae^x \to {y'}_p = Ae^x \to {y''}_p = Ae^x$ we have:
$$Ae^x – 3Ae^x + 2Ae^x = 2e^x$$
$$0 \cdot Ae^x = 2e^x$$
$$0 = 2$$
Which is very bad. What am I doing wrong here?
Best Answer
$y^{\prime \prime}−3y^{\prime}+2y= 2e^x$. Use operator D:
Let $D = d/dx$. So, $(D^2 - 3D + 2)y_p = 2e^x \quad \Rightarrow \quad (D-1)(D-2)y_p = 2e^x \quad \Rightarrow$
$y_p = \dfrac{1}{(D-1)(D-2)}\cdot 2e^x = 2e^x\dfrac{1}{(D+1-1)(D+1-2)}\cdot 1 = 2e^x\dfrac{1}{D(D-1)}\cdot 1 \quad \Rightarrow$
$y_p = 2e^x\dfrac{1}{D(0-1)\cdot 1} = -2e^x\dfrac{1}{D}\cdot 1 = -2xe^x$
Look this blog: http://fatosmatematicos.blogspot.com/2011/02/o-metodo-da-chave-invertida-para-achar.html