1. ## Another DFQ

(dy/dt) - (3/t)y = 4(t^3)cos(5t) , y(pi) = 1

2. Originally Posted by DavidWagner
(dy/dt) - (3/t)y = 4(t^3)cos(5t) , y(pi) = 1
This DE is first order linear and is solved using the integrating factor technique. Read your class notes, textbook or a website like Linear Equations for a refresher on the method.

3. i got

y= (4(t^3)sin(5t))/5 + t^3C

then y(pi) implies C = -.184

Is this correct?

4. Originally Posted by DavidWagner
i got

y= (4(t^3)sin(5t))/5 + t^3C

then y(pi) implies C = -.184

Is this correct?
Can't you check it yourself? I might get hit by a bus tomorrow, then where would you be ..... Empower yourself (and it's good practice for you too):

Substitute $x = \pi$ and check that it gives y = 1.

Substitute the solution into the DE and check that it works.