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Consider an $n^{th}$ order linear differential equation of the form $a_nx^{[n]} +a_{n−1}x^{[n−1]} +···+a_1x′ +a_0x=f(t)$, $(1)$ where the $a_k$ are constants and $x^{[k]}$ denotes the $k^th$ derivative of $x(t)$. The characteristic polynomial associated to this equation is $p(\lambda)=a_n \lambda^n +···+a_1 \lambda +a_0.$ Suppose $f(t) = e^{\alpha t}$, where $\alpha$ is not a root of $p(\lambda)$ ($\alpha$ could be complex). Prove that the ODE above has a special solution of the form $x(t) = ce^{\alpha t}$, for some constant $c \neq 0.$ Use this method to find a special solution to the ODE $x'' −2x′ +3x=e^t$

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What have you tried? –  Artem Dec 1 '12 at 19:16

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The particular case: It is best, I think, to look first at concrete examples. So we look at yours, though it is perhaps too special, since the right-hand side is $e^t$, and therefore $\alpha=1$. Let's not worry about the characteristic polynomial stuff yet, that will come later.

We look for a solution of the DE of the shape $x=ce^t$. Substitute in the left-hand side of the DE. We have $x'=ce^t$ and $y''=ce^t$. So the left-hand side of the DE is $$ce^t-2ce^t+3ce^t.$$ We want this to be identically equal to $e^t$. This will be the case if and only if $c-2c+3c=1$, that is, iff $c=\frac{1}{2}$.


The general case: We look for a constant $c$ such that $x=ce^{\alpha t}$ is a solution of the DE.

Calculate the various derivatives. We have $x=ce^{\alpha t}$. Thus $x'=c\alpha e^{\alpha t}$, and $x''=c\alpha^2 e^{\alpha t}$, and so on. Finally, $x^{[n]}=c\alpha^n e^{\alpha t}$.

Substitute these values in the left side of the DE, and simplify a bit. We get $$(a_0\alpha^n +a_{n-1}\alpha^{n-1}+\cdots +a_n)ce^{\alpha t}.$$ More briefly, this can be written as $$p(\alpha)ce^{\alpha t}.\tag{$1$}$$ We want this to be identically equal to $f(t)$, that is, to $e^{\alpha t}$.

Since by assumption $p(\alpha)\ne 0$, Equation $(1)$ holds iff $$c=\frac{1}{p(\alpha)}.$$ So we have found a $c$ (indeed the only $c$) that works.

Remark: Note that the analysis breaks down if $p(\alpha)=0$. And indeed if $\alpha$ is a root of the characteristic polynomial, then find a particular solution of our DE is more complicated. In that case, no $x$ of the shape $ce^{\alpha t}$ is a solution.

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