Solve ($D^4 + 2D^2 +1)y = x^2 \cos x$ using higher order differential equations

Question

Solve $$(D^4 + 2D^2 +1)y = x^2 \cos x $$ using higher order differential equations

How to decide whether to solve it by Imaginary part or Real part of equations ?

How to solve it by both real as well as imaginary

Answer 1

You could use operator calculus on this one

Solve this by iteration. We have

\begin{align} [D^{(4)}+2D^{(2)}+1]y&=[D''+1][D''+1]y \\ &=[D''+1]Y=x^2 \cos x \end{align} Solve for $y$ then solve $[D''+1]y=Y$.

Answer 2

The characteristic polynomial of the homogeneously equation

$$(D^4 + 2D^2 +1)y =0$$

is given by $p(t)=t^4+2t^2+1=(t^2+1)^2$. $p$ has the zeroes $i$ and $-i$, both of order $2$. Hence the general solution of $(D^4 + 2D^2 +1)y =0$ is given by

$$c_1 \cos x+c_2 x \cos x+c_2 \sin x+c_4 x \sin x,$$

where $c_1,...,c_4 \in \mathbb R.$

A special solution $y_s$ of

$$(*) \quad (D^4 + 2D^2 +1)y =x^2$$

can be found by the "Ansatz" $y_s(x)=ax^2+bx+c.$ Use $(*)$ to derive $y_s(x)=x^2-2.$

Hence the general solution of $(D^4 + 2D^2 +1)y =x^2$ is given by

$$c_1 \cos x+c_2 x \cos x+c_2 \sin x+c_4 x \sin x+ x^2-2,$$

where $c_1,...,c_4 \in \mathbb R.$