How to solve this equation? $ y'' + y = 4xe^x $

Question

I know that $ y'' + y = 4xe^x $ is a sort of unhomogeneous diff equation and that a general solution would include a combo of both the unhomogeneous and the homogeneous solution, and that the geometric and algebraic multitude s, and n play a role here....but I am not certain of their definition nor how to combine it all.

Answer 1

As $1$ is not a root of the characteristic polynomial (of the (homogeneous) differential operator on the left side), the degree of the polynomial factor stays the same and you have to try to fit the parameters in $$ y_p(x)=(ax+b)e^x $$ to the equation.

Answer 2

Hint: $$$$Here is an alternate solution using the Method of Annihilators:$$$$ $y_1=4xe^x$ is clearly annihilated by $D^2-2D+1=(D-1)^2$ where $D$ denotes the derivative operator. Hence, using this annihilator on both sides of the original ODE, the ODE can be "rewritten" as $$(D^2+1)(D-1)^2y=(D+i)(D-i)(D-1)^2y=0$$ $$$$Edit: As pointed out in the comments, after getting the solutions to the equation $(D^2+1)(D-1)^2y=(D+i)(D-i)(D-1)^2y=0$, we need to see which solutions actually satisfy the original equation - the calculations there are going to be exactly like if we used Undetermined Coefficients.

Answer 3

$$y''+y=4xe^x$$

the characteristic equation is $m^2+1=0$ with solutions $m_{12}=\pm\ i$

$$\mathbf{y_{h}(x)=C_1 \cos(x)+C_2\sin(x)}$$

from here you can use Undetermined Coefficients or Variation of parameters \begin{align} y_p(x)&=(Ax+B)e^x \\~ \\ y'_p(x)&=Ae^x+Axe^x+Be^x \\ y''_p(x)&=2Ae^x+Axe^x+Be^x \end{align} replace in the ODE $$ (2Ae^x+Axe^x+Be^x)+(Axe^x+Be^x)=4xe^x \\ (2A+2B)e^x+2Axe^x=4xe^x $$ then $ A=2$ and $ B=-2$ $$\mathbf{y_p(x)=2(x-1)e^x} \\~\\ \mathbf{y(x)=C_1 \cos(x)+C_2\sin(x)+2(x-1)e^x} $$