# Solving the initial value problem of a differential equation

Let $x′′- q(t) x = 0$, $0\le t \lt\infty$ , $x(0)=1$, $x'(0)=1$, where $q(x)$ is monotonically increasing continuous function, then what will be the solution?

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Do you have a typo? You say $q(t)$ in the equation but then say $q(x)$ is monotone. What kind of form for the solution do you expect? I don't think there is a simple general solution considering that even for the simple case $q(t)=t$ one gets the Airy differential equation. –  Alex R. Oct 25 '12 at 20:54
If you have a series expansion $$q\left(t\right) = \sum_{n=0}^\infty a_n t^n$$ with $a_n$ given, you can substitute $$x\left(t\right) = \sum_{n=0}^\infty b_n t^n,$$ where initial conditions give $b_0=b_1=1$, and get $$\sum_{n=0}^\infty \left[\left(n+2\right)\left(n+1\right) b_{n+2} - \sum_{m=0}^{n} a_m b_{n-m} \right]t^{n},$$ or $$b_{n+2} = \frac{1}{\left(n+2\right)\left(n+1\right)}\sum_{m=0}^{n} a_m b_{n-m}.$$