How can I find particular solution to:
y" + y = (cos x)^2

Question

How can I find particular solution to:
y" + y = (cos x)^2

amistre64 · Answer

a few ways, but the "wronskian" method might be simplest

amistre64 · Answer

y = yh + yp

yh is the homogenous solution suc that:

yh = c1y1 + c2y2

then yp can be set up as:

W1      W         W2
y1       0          y2
y'1  (cos x)^2  y'2

cross those

anonymous · Answer

I've used (cos x)^2 = (1 + cos2x)/2
Therefore getting: A + Bcos2x + Csin2x
Let me try this way, I haven't learnt Wronskian yet

amistre64 · Answer

ah, so you rewrote to use the "table"

amistre64 · Answer

make sure your not using up one of your homogenous parts as a double

anonymous · Answer

sorry, what do you mean by using it as a double?

amistre64 · Answer

for he solution to be linearly independant; you need to make sure you dont use the same parts more than once

amistre64 · Answer

for example,if youve already got a c1 cos(2x) then you will need to introduce an extra variable into it such as; x cos(2x)

anonymous · Answer

Ahh I see. No the complimentary part is: c1cos(x) + c2sin(x). So I don't think that'll be an issue.

amistre64 · Answer

good :)

anonymous · Answer

I think I got the answer. Thanks for the help :)

amistre64 · Answer

yw