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[1] What is the largest open interval for which a unique solution of the initial value problem $$ty' + \frac{1}{t+1}y = \frac{t-2}{t-3}$$ $$y(1) = 0$$ is guaranteed.

  • by observation, I'd think that the answer to this would be from -1 to 3 from looking at the $t+1$ and $t-3$ term.
  • The answer is $ 0 < t < 3 $, so I am quite confused.

Similarly (but unrelated)

[2] For what a is the solution of the Initial Value Problem $y ′ − y + 2e^{-t} = 0$, y (0) = a, on interval t ≥ 0 ?

  • I attempted this problem by first solving it
  • Resulting in $y = 2e^{-t} + Ce^{t}$ but I guess thats not very helpful.

What condition should I be looking for when solving these type of problem?

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Am I missing something, or are [1] and [2] essentially unrelated questions? Please split them, e.g., by removing [2] from here and making [2] a separate question. – user53153 Feb 18 '13 at 4:14
up vote 2 down vote accepted

1) In general, this kind of question requires this:

In this case, the equation is linear, so you just have to check where the coefficients are continuous, basically.

First, you need to divide by $t$ to put it in the form $$ y'+a(t)y=b(t). $$ Your coefficients $a(t)$ and $b(t)$ are indeed continuous on $(0,3)$ and $1\in(0,3)$. So there exists a unique solution to your IVP on $(0,3)$. This can't be done on a larger interval, because of the $t$ and the $t-3$ at the denominator.

2) You are there. You solution is bounded on $[0,+\infty)$ if and only if $C=0$. Now $y(0)=a=2$.

Answer: $a=2$.

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To solve [1] you would naturally want to apply a technique like integrating factors, however that would require you to divide by $t$. This is a simple explanation why you would further restrict your interval of validity of your solution.

For [2], you need to figure out what $C$ values will give you a bounded function. Clearly 2$e^{-t}$ is bounded however for any $C\ne0$ the second half of your result will grow as you go to infinity so you need $C$ to be zero. Just choose $a$ so that it forces $C$ to be zero.

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