How do I solve #y'' + y = -2 sin(x)# with the initial conditions #y(0) = 0# and #y'(0) = 1#?

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1 Answer
Gió
Mar 16, 2015

I am not sure how much you know about 2nd order differential equation but I'll try to be as clear as possible. (try the theory in K. A. Stroud, "Engineering Mathematics", it is very good!)
Your solution is made of two parts:
General Solution=Complementary Function+Particular Integral=CF+PI
CF is given equating to zero your equation and transforming it into a 2nd degree equation introducing an auxiliary factor #m#.
PI is given depending upon the function #f(x)# you have after the #=# sign.
So you have:
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Hope it helps

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