An oddly puzzling differential equation

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SUMMARY

The differential equation t*y'(t) + 4*y = 0 with the initial condition y(3) = 2 has the solution y(t) = 162t^(-4). The discussion reveals that a common mistake in solving this equation involves misapplying exponent and logarithm rules. Specifically, the correct transformation from ln(y) = -4ln(t) should yield y = e^(-4ln(t)) = t^(-4)e^(-4), leading to the final solution. The constant of integration is crucial and should not be overlooked in the solution process.

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Homework Statement


Solve t*y'(t) + 4*y = 0; y(3) = 2. This is the solution---> Ans: y(t) = 162t^(-4)

Homework Equations


I need to know how my professor got this answer.

The Attempt at a Solution


I attempted by subtracting 4y to the other side and separated the variables to yield
[tex]\int(dy/y)[/tex] = -4[tex]\int(dt/t)[/tex]
for which you get ln(y) = -4ln(t)
=> y = t*e^(-4).
Can anyone help me on this?
(I left out the constant of integration since that only pertains to part 2 for a particular solution)
 
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Are you sure you applied your exponent/logarithm rules correctly?

By the way, the constant of integration is always relevant. Forgetting it is in the same class of mistakes as forgetting that -2 is a solution to x²=4.
 
have a look at your exponent step

ln(y) = -4ln(t)
now take exponential
eln(y) = y = e-4ln(t)

note - e-4.eln(t)= e-4 + ln(t), not what you have...
 

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