3rd Order Differential Equation

Question

Find a third order homogenous linear equation which has a particular solution $$y_p= e^x - 2e^{-2x} + 5xe^{-2x}$$ I tried solving it direct, but couldn't . Seems a reserve would actually do but I don't know how to get there. Pls I need help

Answer 1

I use the 'D' operator from Piaggio's "an Elementary Treatise on Differential Equations", an axiomatic and justified approach to treating $\frac{d^{n}}{dx}$ as an algebraic quantity, $D^{n}$

Accordingly, your problem reduces to the following: Find A, B, C, E and F, such that:
$(AD^{3} + BD^{2} + CD + E)y = F(x)$

Clearly due to general solutions of equations of this nature being in the form
$y = (a + bx + cx^{2} + ...)e^{\alpha x} + (a' + b'x + c'x^{2} ...)e^{\beta x} +...$
Working backwards, $e^{x} - 2e^{-2x} + 5xe^{-2x}$ can be written $(1)e^{x} + (-2 + 5x)e^{-2x}$

To clarify, $\alpha, \beta...$ are roots of the characteristic equation, and it can be shown with a little effort that the degree of each polynomial coefficient to each exponential term is precisely equal to the repeatedness of each root.

Therefore the characteristic equation is $(D+2)^{2}(D-1)$, and expanding gives you your values of $A, B, C, E$

It follows that, since your particular solution is in the form of the general solution, that $F(x) = 0$, a common result.