The solution to the Differential Equation

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SUMMARY

The discussion focuses on solving the initial value problem for the differential equation y' - (3/t)y = 0 with the initial condition y(1) = -10. The integrating factor was correctly identified as u(t) = t^-3, which is derived from the integral of -3/t. The solution process involves substituting this integrating factor back into the equation, leading to the general solution y = -Ct/3. The confusion arises regarding the presence of t^3 in the potential answers, indicating a misunderstanding in the integration step.

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  • Understanding of first-order linear differential equations
  • Knowledge of integrating factors in differential equations
  • Familiarity with integration techniques, specifically integrating 1/t
  • Basic algebraic manipulation skills
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  • Study the integration of functions involving logarithmic forms, particularly 1/t
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Northbysouth
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Homework Statement


The Initial value problem:

y' - (3/t)y = 0

y(1) = -10

Has the solution:

I have attached an image of the question


Homework Equations





The Attempt at a Solution



Firstly I found the integrating factor:

u(t) = e∫-3/t dt

u(t) = -3/t

I plugged this back into the original equation:

(-3/t)y' + (9/t2y = 0

∫[-(3/t)y]' dt = ∫ 0 dt

(-3/t)y = C

y = -Ct/3

But I don't understand where I go from here. The possible answers all include t3 and I don't understand where that has come from. What step did I miss?
 

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Northbysouth said:

Homework Statement


The Initial value problem:

y' - (3/t)y = 0

y(1) = -10

Has the solution:

I have attached an image of the question


Homework Equations





The Attempt at a Solution



Firstly I found the integrating factor:

u(t) = e∫-3/t dt

u(t) = -3/t

What is the integration of 1/t?
 
Yes, I see my mistake now.

u(t) = t^-3
 

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