Solving a First Order Differential Equation with Initial Conditions.

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SUMMARY

The forum discussion centers on solving the first-order linear differential equation given by t(dy/dt) + 8y = t^3 with the initial condition y(1) = 0. The user correctly rearranged the equation to dy/dt + 8y/t = t^2 and applied the integrating factor e^(∫8/tdt) = t^8. After integrating, the user derived the solution y = (t^3)/11 - 1/(11t^8) but questioned its accuracy. Other participants confirmed the solution is correct, suggesting a possible typo in the reference material.

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  • Understanding of first-order linear differential equations
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  • Familiarity with initial value problems
  • Basic calculus, specifically integration techniques
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  • Practice solving initial value problems with varying conditions
  • Explore common errors in solving differential equations
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uber_kim
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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!
 
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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!
 

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