Question

Suppose that a fourth order differential equation has a solution y = -9e^(4x)xsin(x) Find such a differential equation, assuming it is homogeneous and has constant coefficients.

Answer 1

So you're looking for something that looks like this:
\[ay^{(4)}+by'''+cy''+dy'+ey=0\]
(Here, \(e\) is a constant. I'll differentiate between this \(e\) and the exponential function \(e^x\) by writing the latter as \(e^x=\exp{x}\).)

The characteristic equation is
\[ar^4+br^3+cr^2+dr+e=0\]
You know that one solution is \(y=-9\exp{(4x)}x\sin x\). Think of the constant more generally: write \(C_1=-9\). Since the solution is the product of an exponential and a trig function, \(C_1\exp(\lambda x)\sin(\mu x)\), you know that the characteristic equation has complex roots of the form \(\lambda\pm\mu i\).

Also, because of the power of \(x\) in the solution, you know that the root is repeated. The multiplicity of the root can be inferred from the degree of the differential equation.

The complex roots, judging by the given solution, must then be \(4\pm i\), with multiplicity 2. Find a characteristic equation with these roots, and you'll be set.