How to obtain a particular solution for this linear ODE?

Question

How to obtain a particular solution for this linear ODE? $$y'' + 2y' - 8y = 2e^{-2x} - e^{-x}.$$

I am still struggling with this kind of equations, and we didn't see an example in class like this. How would i use the annihilator method? I already got homogeneous solutions, but i cant seem to get and idea to obtain a particular solution.

Answer 1

Hint. Find a particular solution $y_1$ and $y_2$ for each of these differential equations: $$1)\;\; y'' + 2y' - 8y = e^{-2x}\qquad,\qquad 2)\;\; y'' + 2y' - 8y = e^{-x}.$$ Then, by linearity, $2y_1-y_2$ is a particular solution of the given differential equation.

Since the characteristic polynomial is $z^2+2z-8=(z+4)(z-2)$, it follows that $y_1(x)=Ae^{-2x}$ and $y_2(x)=Be^{-x}$ where $A$ and $B$ are real constants to be found.

Answer 2

Hint:

You can handle both terms separately (by the superposition principle).

Then for the first term, try a solution of the form $e^{-2x}$. You have

$$(e^{-2x})''+2(e^{-2x})'-8e^{-2x}=-8e^{-2x}$$ which is what you need, to a multiplicative coefficient.

You should be able to continue.

Answer 3

With annihilator method $$y'' + 2y' - 8y = 2e^{-2x} - e^{-x}.$$ $$(D^2 + 2D - 8)y = 2e^{-2x} - e^{-x}.$$ $$(D^2 + 2D - 8)(D+2)y = - e^{-x}.$$ $$(D^2 + 2D - 8)(D+2)(D+1)y =0$$ Since $(D^2 + 2D - 8)=(D-2)(D+4)$ $$(D-2)(D+4))(D+2)(D+1)y =0$$ $$ \implies y=\color{blue}{y_h}+y_p$$ $$\implies y(x)= \color{blue}{c_1e^{2x}+c_2e^{-4x}}+\underbrace{c_3e^{-2x}+c_4e^{-x}}_{\text {particular solution }y_p}$$ Plug $y_p$ in the equation to get $c_3$ and $c_4$