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?
Solving the initial value problem of a differential equation
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ordinary-differential-equations
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0http://eqworld.ipmnet.ru/en/methods/methods-ode/Khorasani2003.pdf may help you. – 2012-10-25
1 Answers
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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}. $