Thread: 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.

You'll notice the first group of terms cancel with the last group leaving



SO, can you choose C and D so that the left side = the right side?

yeah it makes sense. I worked out d=15 and c=0. hence yp= 15tsin(5t)

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