An oddly puzzling differential equation

iatnogpitw
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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
\int(dy/y) = -4\int(dt/t)
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...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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