# Easiest way to solve $F^{(n)}(y) + F(y) = c$

I'm supposing I have the following $n$th order differential equation:

$$F^{(n)}(y) + F(y) = c$$

Say that I have $F(0)$, $F'(0)$, etc., and $c$. What is the easiest way to solve the equation? I'm really wondering if there are shortcuts to solving simple, higher order differential equations. The equation above is an excellent example of what I'll be working with most.

Also: Note that you may assume that $c$ takes on an ideal value if it helps.

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Find roots $\lambda_i$ of $x^n+x=0$, devise the general solution $F(y)=c+\sum a_i\exp(\lambda_iy)$, solve for coefficients using initial conditions and linear algebra on the resulting system. There's a more general technique to transform $n$-th order linear (in)homogeneous differential equations in $u$ into systems of $n$ first-order linear (in)homogeneous DEs by writing $u_n=d^{n-1}u/dt^{n-1}$ for $n=1,2,\dots$ and then obtaining $\vec{u}'=A(t)\vec{u}+\vec{b}(t)$. – anon Sep 9 '11 at 2:52
Isn't it the standard constant coefficient linear ODE? Are there any easier ways than the standard "characteristic equations" method? – Jack Sep 9 '11 at 3:55
Woops, that should be $x^n+1=0$ in my above comment. Duh. – anon Sep 9 '11 at 5:30

First, the homogeneous (i.e. "$=0$" instead of "$=c$") equation: $$(D^n+I)F(y) = 0$$ The roots of the polynomial $D^n+1$ are, $e^{\pi i/n}$, $e^{\pm3\pi i/n}$, $e^{\pm5\pi i/n},$ $\ldots,$ $e^{\pm\ell\pi i/n}.$ In other words $e^{k\pi i/n}$ for $k=1,\ldots,\ell$ where $\ell$ is the largest odd number $\le n/2$. So $$y = \sum_k c_k\exp\left({e^{\pm k\pi i/n}t}\right)$$ are the solutions, where the coefficients $c_k$ are complex numbers.

But what about "$=c$"? Find a polynomial solution. Add that to the solution of the homogeneous equation.

Then find the coefficients that make the whole thing satisfy the initial conditions.

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I don't know what your $F$ is (linear or not etc.) but, I would proceed with rewriting it as

$$\begin{pmatrix} F'(y)\\ F''(y)\\ \vdots\\ F^{(n)}(y)\\ \end{pmatrix} = \begin{pmatrix} 0 &1& &\ldots &0\\ 0 &0 &1 &\ldots &0\\ \vdots&&&\ddots\\ 0&\ldots&0&0&1\\ -1 &0 &0 &0&0 \end{pmatrix} \begin{pmatrix} F(y)\\ F'(y)\\ \vdots\\ F^{(n-1)}(y)\\ \end{pmatrix}+ \begin{pmatrix} 0\\ 0\\ \vdots\\ 0\\ c\\ \end{pmatrix}$$ and integrate.

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F is linear.... – Matt Groff Sep 9 '11 at 3:06
@Matt: I don't think $F$ has to be linear. The linearity is in the differential operator instead of the function $F$. – Jack Sep 9 '11 at 4:02