Linear Diff. Eq. & Uniqueness

[calvert11]

1. The problem statement, all variables and given/known data
Solve the IVP. Is your solution unique? Explain.

ty' + (t-2)y = (t^4)*(e^t)

y(0)=0

2. Relevant equations

Theorem:

If p(t) and g(t) are continuous functions on an open interval a< t < b and the interval contains t0, then there is a unique solution to the IVP on that interval.

3. The attempt at a solution

I'm rather sure I've solved the equation correctly, but I end up with something like this:

y = \frac{t^2}{e^t} X [....etc.]

Now if I input the initial values, I get 0 = 0.

Furthermore, since p(t) = (t-2)/t where t cannot = 0, the Theorem would fail as no interval exists that satisfies it.

So, I'm a bit confused about how to answer this question. Assuming I solved the equation correctly, is 0 = 0 a solution? And since the equation fails the Theorem does that mean it is not unique? I do hope I'm making sense.

[HallsofIvy]

1. The problem statement, all variables and given/known data
Solve the IVP. Is your solution unique? Explain.

ty' + (t-2)y = (t^4)*(e^t)

y(0)=0

2. Relevant equations

Theorem:

If p(t) and g(t) are continuous functions on an open interval a< t < b and the interval contains t0, then there is a unique solution to the IVP on that interval.

3. The attempt at a solution

I'm rather sure I've solved the equation correctly, but I end up with something like this:

y = \frac{t^2}{e^t} X [....etc.]
This is certainly NOT a solution to the equation. It is an integrating factor for the equation. Is that what you meant?

Now if I input the initial values, I get 0 = 0.

Furthermore, since p(t) = (t-2)/t where t cannot = 0, the Theorem would fail as no interval exists that satisfies it.

So, I'm a bit confused about how to answer this question. Assuming I solved the equation correctly, is 0 = 0 a solution?[/quote]
No, "0= 0" is not a 'solution', it isn't even a function. It is true that y(t)= 0, for all t, is a solution.

And since the equation fails the Theorem does that mean it is not unique? I do hope I'm making sense.
That means it may not be unique. Can you find another function satisfying that differential equation and initial value? Hint: Can you solve that differential equation with y(1)=0. Now, what if you took y(t)= 0 for t< 1 and y(t)= that solution for y> 1?

[calvert11]

This is certainly NOT a solution to the equation. It is an integrating factor for the equation. Is that what you meant?

No, I found the integrating factor to be \frac{e^t}{t^2}


No, "0= 0" is not a 'solution', it isn't even a function.
lol yeah. that was a silly question. So would y(0)=0 have no solution then?

Sorry, I'm really unaccustomed to using the layout tools here so please bare with me. This is "in general" what I came up with before putting in the initial values y(0)=0

y = (t^2/e^t)*[....etc. + C] where C is some constant. Obviously 0^2 = 0 so everything in the brackets becomes 0 and since y = 0 then 0 = 0.


That means it may not be unique. Can you find another function satisfying that differential equation and initial value? Hint: Can you solve that differential equation with y(1)=0. Now, what if you took y(t)= 0 for t< 1 and y(t)= that solution for y> 1?
let me try that. but would you mind giving some input on what I've written above?