Question

Suppose that a fourth order differential equation has a solution y=3e^(4x)xcos(x). (a) Find such a differential equation, assuming it is homogeneous and has constant coefficients.

Answer 1

differentiate this up to 4 times and make some expression with it.

Answer 2

That's one way of doing it.

Answer 3

still confusing!

Answer 4

So you have to find a 4th order ODE that looks like this:
\[ay^{(4)}+by'''+cy''+dy'+ey=0\]
The characteristic equation is then
\[ar^4+br^3+cr^2+dr+e=0\]
You know one solution to the ODE is \(y=3e^{-4x}x\cos x\). This gives you a few bits of information:
∙ \(e^{\lambda x}\cos \mu x~\Rightarrow~r=\lambda+\mu i\) is a solution to the characteristic equation. (Since complex roots come in conjugate pairs, you also know that \(r=\lambda-\mu i\) is also a solution to the characteristic equation.)
∙ The power of \(x\) in the solution indicates a repeated root.

So, you have to find \(a,b,c,d,e\) such that
\[ar^4+br^3+cr^2+dr+e=0\]
has roots \(r=-4+i\), with multiplicity 2.