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What is the general solution of the differential equation of the form $y^{(4)} + ay = f(x)$, where $f(x)$ is a polynomial of $x$.

In my textbook, I have found the method of finding the general solution of high order differential equations of the form y^{(n)} = g(y^{(n-1)}, \cdots, y', x) or Euler equations. But this equation is of neither kind.

It is not difficult to find a particular solution. For example, when $a \neq 0$ and $f(x)$ is of degree $2$, $f(x) = b_2x^2 + b_1 x + b_0$. Then $y = \frac{1}{a}(b_2x^2 + b_1 x + b_0 + ae ^{\sqrt[4]{-a}x})$ is a particular solution.

But what will the general solution be like?

I am doing algebra everyday, so I know little about differential equations. Thanks to everyone for viewing or help.

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    Another remark. Your solution fails when $a=0$. Figure out why!2011-10-14

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Note that General solution (G.S.)=Complementary function (C.F.)+Particular Integral (P.I). It looks your main problem is with the C.F. part. From your given data, I assume $a\ne0$ (the case $a=0$ is very simple. If you can't figure out, just let me know it).

For C.F.:

Case1: a>0

The roots of $m^4+a=0$ are $\pm a^{\frac{1}{4}}[\frac{1\pm i}{\sqrt{2}}]$ which are two complex conjugate pair: $b\pm bi$ and $-b\pm bi$ where for brevity we put $b=\frac{a^{\frac{1}{4}}}{\sqrt{2}}$.

Hence the C.F. is $y_{cf}=e^{bx} (A_1\cos bx+A_2\sin bx)+e^{-bx} (A_3\cos bx+A_4\sin bx)$

Case2: $a<0$ The roots of $m^4+a=0$ are $\pm (-a)^{\frac{1}{4}}$ and $\pm (-a)^{\frac{1}{4}}i$. So noting that there are two distinct real roots $\pm c$ one complex conjugate pair $\pm ci$ (for brevity we put $c=(-a)^{\frac{1}{4}})$, the C.F is $y_{cf}=A_1e^{cx}+A_2e^{-cx}+ A_3\cos cx+A_4\sin cx$. In both cases, $A_i$'s are arbitrary constatnts.

For the P.I.:

$y_{pi}=\frac{1}{D^4+a} f(x)=\frac{1}{a(1+\frac{D^4}{a})} f(x)=\frac{1}{a}(1+\frac{D^4}{a})^{-1} f(x)=\frac{1}{a}[1-\frac{D^4}{a}+\dotsc]f(x)=\frac{f(x)}{a}$, since your $f(x)$ is a polynomial of degree less than 4.

Combining, write down the G.S. as $y=y_{cf}+y_{pi}$.