Solving a First Order Differential Equation with Initial Conditions.

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
uber_kim
Messages
7
Reaction score
0

Homework Statement



Solve the initial value problem:

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

Homework Equations



None?

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!
 
Physics news on Phys.org
uber_kim said:

Homework Statement



Solve the initial value problem:

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

Homework Equations



None?

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?

Your answer is correct. Are you comparing with the book's answer? As long as you are working the same problem, your answer checks.
 
Hmm, strange. Maybe a typo. Thanks!