Differential Equations and Initial Values

[Northbysouth]

Homework Statement

If y is the solution of the initial value problem

y'(t) + y(t)/t = 1/t⁴

y(2)=1

What is y(3)

Homework Equations

The Attempt at a Solution

I u(t) = e^{∫1/t dt} = e^{ln 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.

 

[mfb]

I u(t) = e^{∫1/t dt} = e^{ln t} = t

∫[yt]' dt = t^-4 dt

yt = -1/3t³ + C

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.

 

[LCKurtz]

Homework Statement

If y is the solution of the initial value problem

y'(t) + y(t)/t = 1/t⁴

y(2)=1

What is y(3)

Homework Equations

The Attempt at a Solution

I u(t) = e^{∫1/t dt} = e^{ln t} = t

∫[yt]' dt = t^-4 dt

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

 

[HallsofIvy]

I have no idea what you are doing here.

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

Let's check:
I think "yt" should be y(t)?

No, it clearly should not be: integrating (ty)' gives t times y.

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.