# Solving a First Order Differential Equation with Initial Conditions.

1. Jan 16, 2013

### uber_kim

1. The problem statement, all variables and given/known data

Solve the initial value problem:

t(dy/dt)+8y=t^3 where t>0 and y(1)=0

2. Relevant equations

None?

3. The attempt at a solution

It's a linear equation, so rearranged to dy/dt+8y/t=t^2.

Took the integrating factor e^(∫8/tdt)=t^8 and multiplied through.

(t^8)dy/dt+8t^7y=t^10.

∫(t^8y)'dt=∫t^10dt
t^8y=(t^11)/11 + C

Solve for C, I got -1/11.

Final solution is y=(t^3)/11 - 1/(11t^8)

This isn't right, though. Does anyone see where I made the mistake?

Thanks!

2. Jan 17, 2013