2° order linear, homogeneous, variable coefficients

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sin(x) * y''(x) + 2cos(x) * y(x) = 0
y(0) = 0
y'(0) = 1

how do I solve it?
(I know the solution because I have created the diff. equation starting from a simple function).

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[Simon Bridge]

This is equation of form: $y'' + p(x)y' + q(x)y = g(x)$ where $p=g=0$.
Have you tried looking it up? Do you know one solution to the DE (then try reduction of order)?

 

[lightarrow]

This is equation of form: $y'' + p(x)y' + q(x)y = g(x)$ where $p=g=0$.
Have you tried looking it up? Do you know one solution to the DE (then try reduction of order)?

Yes I tried but as you suggest I've only found a text on DE where it says "...if you know two independent solutions of the equation..."
Apart from the fact that I only know one solution, which is e^-x⋅sin(x), I was asking if there were a way to find such solution/s without having to try by chance (in this case I didn't because, as I wrote, I started from that function and then I found a DE that it satisfies).

I also tried with the substitution y(x) = e^t(x) but the new DE in t(x), that is: t''(x) + [t'(x)]² = -2cot(x), is a Riccati equation (after substitution t'(x) = u(x) ) the resolution of which becomes again the initial DE .

I'm wondering if we could do anything by Fourier transforming the equation...

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[lightarrow]

Yes I tried but as you suggest I've only found a text on DE where it says "...if you know two independent solutions of the equation..."

I correct myself: actually we only have to know one, y1(x). To find the general solution y(x) we write y(x) = y1(x)*z(x) and after substitution in the DE we find another DE in z(x) which in this case is possible to solve, but not in terms of finite combination of elementary functions, it's infact the primitive of the function e^2x / [sin(x)]^2.

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[Simon Bridge]

Well done. The process is called "reduction of order".
For those who google here later and don't know what I'm talking about:
http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx

... note: I have not checked your answer.