This may help with your integrals
and
so
the other
You can use the reduction formula for u2 or the half angle formula
I hope this helps
Hey all,
heres the differential equation to be solved using Variation of parameter's:
y" + y = sin(x)^2
i got this far:
u1'cos(2x) + u2'sin(2x) =0
-2*u1'*sin(2x) + 2*u2'*cos(2x) = sin(x)^2
solving for u1' gives: (-sin(x)^2*sin(2x))/2
and u2': (-sin(x)^2*cos(2x))/2
The final solution should take the form
u1*cos(2x) + u2*sin(2x)
so whats left is integrating u1' and u2'. Unfortunately due to problems with integrating and simplfying trig expressions i'm not getting the solution which is
1/8(1-xsin(2x))
thanks alot
Alright, lets try this again...
If this is the case then our complementry solution is
so computing the wronskian
then
and
this would give the integrals
and
by the double angle formula...
using the double angle formula again on sine...
finally, now we let and
This yeilds...
is a bit better
rewritting with the double angle formula
with and we get
I hope this gets it...
I follow all of your work,
but it seems when i try to get my final particular solution in the form
yp = u1*sin(2x) + u2*cos(2x)
i'm not getting the solution
1/8(1-x*sin(2x))
am i wrong in saying the particular solution yp should be in the form
yp = u1*sin(2x) + u2*cos(2x)