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

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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First of all: if $a > 0$, $\sqrt[4]{-a}$ isn't defined. Next (as pedja commented): think of finding the homogenous solution first. Start by finding all the solutions to $y^{(4)} + a y = 0$ (consider both signs!). Then, for the particular solution, you can forget about what you did before. I suggest you look at polynomial solutions. Since $f(x) = \sum a_k x^k$ for a finite number of $a_k$, it makes sense to use the Ansatz $y(x) = \sum b_k x^k$. Plug this into your equation and find a relation between the $b_k$ and the $a_k$. – Gerben Oct 14 '11 at 10:23
@pedja: Thank you very much. Now I know I should add the general solution of the homogeneous equation to the particular solution. But, the "[previous section][1]" only gives the solution to homogeneous equation of order $2$, and I find it hard to extend the method to higher order... What can I do? Thanks again. [1]: sosmath.com/diffeq/second/constantcof/constantcof.html – ShinyaSakai Oct 14 '11 at 10:31
@Gerben: Thank you very much. Here $\sqrt[4]{-a}$ can denote any $b$ of the four numbers such that $b^{4} = - a$. This is possible to be defined. Maybe I should explain this in my statement of the question. – ShinyaSakai Oct 14 '11 at 10:35

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 $y=y_{cf}+y_{pi}$.

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