# Differential Equations and Initial Values

1. Feb 14, 2013

### Northbysouth

1. The problem statement, all variables and given/known data
If y is the solution of the initial value problem

y'(t) + y(t)/t = 1/t4

y(2)=1

What is y(3)

2. Relevant equations

3. The attempt at a solution

I u(t) = e∫1/t dt = eln t = t

∫[yt]' dt = t-4 dt

yt = -1/3t3 + C

Plugging in my initial value I solve for C:

1 = -1/3(24) + C/t

C = 2 + 1/3(23)

C = 2 + 1/24 = 49/24

y(3) = -1/3(34) + 49/24

y(3) = 3961/1944

The answer should be 149/216 but I can't see where I'm going wrong. Help is appreciated.

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2. Feb 14, 2013

### Staff: Mentor

I have no idea what you are doing here.

Let's check:
I think "yt" should be y(t)? And the t3 is part of the denominator?
Like this: y(t) = -1/(3t3) + C
y'=1/t4
Therefore, $\frac{1}{t^4} -\frac{1}{3t^4} + \frac{C}{t} = \frac{1}{t^4}$
This cannot be true for all t.

3. Feb 14, 2013

### LCKurtz

That step is incorrect. First write the correct equation $(ty)' = ?$, then integrate it.

4. Feb 15, 2013

### HallsofIvy

Staff Emeritus
He's finding the "integrating factor" for a linear differential equation. But, as LCKurtz says, he used it improperly. Multiplying both sides of y'(t) + y(t)/t = 1/t4 by t gives
ty'(t)+ y(t)= (yt)'= 1/t3
No, it clearly should not be: integrating (ty)' gives t times y.