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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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bdas: Welcome to math.SE. since you are a new user, we wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Further, it would be better if you could typeset your problem so that it is easy for people to read. Kindly look… for more details. – user17762 Oct 25 '12 at 20:41
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
up vote 2 down vote accepted

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 $$ b_{n+2} = \frac{1}{\left(n+2\right)\left(n+1\right)}\sum_{m=0}^{n} a_m b_{n-m}. $$

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