I'm looking for a solution (general expression) for the following second order ODE: \begin{equation} \frac{d^{2} x}{dt^{2}}-f(t)\frac{d x}{dt}-2 f(t) g(t)=0 \end{equation}
where $f(t),g(t)$ are continuous and differentiable functions.
Second order ode with time dependent coefficients $\frac{d^{2} x}{dt^{2}}-f(t)\frac{d x}{dt}-2 f(t) g(t)=0$
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calculus
analysis
ordinary-differential-equations
functional-equations
1 Answers
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$$ x'' -f(t) \left[x' -2g(t)\right] = 0 $$ let $v = x' -2g(t)$ then we have $$ v' +2g' = x'' $$ or $$ v' +2g' -f(t) v = 0 $$ so a general solution could look like $$ v(t)\mathrm{e}^{\int_0^t f(s)ds} = \int_0^t 2g'(s)\mathrm{e}^{\int_0^s f(\tau)d\tau} ds $$ depending on if you can solve the integrals ... $\textbf{edit:}$ $$ v(t) = \mathrm{e}^{-\int_0^t f(s)ds}\int_0^t 2g'(s)\mathrm{e}^{\int_0^s f(\tau)d\tau} ds = x' - 2g(t) $$ so we have the final integral being $$ x(t) = \int_0^t \left[2g(\tau) + \mathrm{e}^{-\int_0^\tau f(s)ds}\int_0^\tau 2g'(s)\mathrm{e}^{\int_0^s f(s')ds'} ds'\right]d\tau $$