# General solution question

• Jul 15th 2010, 03:51 AM
alpha
General solution question
Find the solution of y'' + 25y = 150cos(5t)

I have found the the complimentary function to be y= Acos(5t) + Bsin(5t)

When I worked out the particular integral I came as far as this:

t(-25C cos(5t) - 25Dsin(5t)) - 10Csin(5t) + 10Dcos(5t) + 25t(Ccos(5t) + Dsin(5t)) = 150 cos (5t)

What i dont unerstand is why the above simpifies to 150cos(5t)??

Thank you for any help.
• Jul 15th 2010, 04:11 AM
Jester
You'll notice the first group of terms cancel with the last group leaving

$\displaystyle -10C\sin(5t) + 10D\cos(5t) = 150 \cos (5t)$

SO, can you choose C and D so that the left side = the right side?
• Jul 15th 2010, 04:19 AM
alpha
yeah it makes sense. I worked out d=15 and c=0. hence yp= 15tsin(5t)
• Jul 15th 2010, 04:45 AM
CaptainBlack
Quote:

Originally Posted by alpha
Find the solution of y'' + 25y = 150cos(5t)

I have found the the complimentary function to be y= Acos(5t) + Bsin(5t)

When I worked out the particular integral I came as far as this:

t(-25C cos(5t) - 25Dsin(5t)) - 10Csin(5t) + 10Dcos(5t) + 25t(Ccos(5t) + Dsin(5t)) = 150 cos (5t)

What i dont unerstand is why the above simpifies to 150cos(5t)??

Thank you for any help.

If you have a table of Laplace transforms available you can do this by taking the LT of the ODE with initial conditions y(0)=y'(0)=0, Simplify and rearrange and look up what the result is the LT of.

CB