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When the coefficients are not constant, and one solutions is known, it is easy to use reduction of order to compute the second solution. But what if both solutions are unknown, is there any general approach to the solutions other than guessing one of the solutions? For example: $y''+sin(t)y'+cos(t)y=0$

How could one solve this equation if there is no solution provided ?

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Jonas answered the general question.

Concerning the specific case you give as an example $$y''+\sin(t)\,y'+\cos(t)\,y=0$$ the equation simplifies if we start setting $y=e^z$ which makes the equation to be $$z''+z' \left(z'+\sin (t)\right)+\cos (t)=0$$ for which an "obvious" solution seems to be $z=\cos(t)+c_1$. So $y=c_2\, e^{\cos(t)}$ is a solution.

Is there any other ? This is a question.

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  • $\begingroup$ Very nice solution ! $\endgroup$ – superman Mar 3 '16 at 5:59
  • $\begingroup$ Thank you ! The problem is that I am not sure at all that it is a nice solution !! Cheers. $\endgroup$ – Claude Leibovici Mar 3 '16 at 6:02
  • $\begingroup$ If you know one solution $y_0$, then you can find another by setting $y=y_0 z$ and deriving a first-order ODE for $z'$. If I haven't made any mistakes, another solution (linearly independent of the first one) is $y(t)=e^{\cos t} \int_0^t e^{-\cos s} ds$. $\endgroup$ – Hans Lundmark Mar 25 '18 at 15:09
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There is no general method to solve a linear equation with nonconstant coefficients.

However, since in the particular example that you give the coefficients are periodic we do know the structure of the solutions, in view of Floquet theory. Still this is often of little use since there is also no general method to compute the characteristic exponents of the nonperiodic part, which govern the asymptotic behavior of the solutions.

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