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here's my question

Consider the differential equation: $t\frac{dg}{dt} = 2g.$

I got that the general solution is $g = ct^2$. However, I don't understand how to answer these questions:

  • What is the unique solution for the IVP $g(-1) = -1$ on the interval $-2 < t < 0$. What is the largest interval on which the solution is unique?

I think the answer is $g=-t^2$. Then, that would mean that the maximum interval is $-\infty < t <0$. Right?

  • Using the same IVP, what are three possible solutions for $-2 < t <2$
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up vote 2 down vote accepted

Looks like your answer is right. For the very last part, your original solution (call it $f$) certainly works in $-2 < t < 2$. Since the solution has to satisfy the initial condition, so it must be $f$ on $-2 < x \leq 0$. At $0$, it can switch to a different function, i.e. any solution which in general has the form $ct^2$. So the following is the general solution and you can get three specific one by choosing three specific constants $c$. $$ h(x) = \begin{cases} f(x) & -2 < x \leq 0 \\ ct^2 & 0 \leq x < 2 \end{cases} $$

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I am not in total agreement. It seems to me that the solutions should be $$h(x) = \cases{f(x)&if $-2<x\le0$\cr ct^2&if $0\le x<2$}$$ where $c$ can be any real number. – Stephen Montgomery-Smith Nov 30 '13 at 5:56
@StephenMontgomery-Smith I agree your solution is more general, but my definition of $h$ still works (special case of your solution with $c = 0$). The last question only asks for three possible solutions anyway. – Pratyush Sarkar Nov 30 '13 at 15:48
Yes, but $g$ isn't a solution. It doesn't satisfy the initial conditions. – Stephen Montgomery-Smith Nov 30 '13 at 15:54
@StephenMontgomery-Smith Oh right, that's true. I'll fix that. But $h$ still satisfies the initial conditions. – Pratyush Sarkar Nov 30 '13 at 16:01
@StephenMontgomery-Smith Thanks for the correction. I think I have it correct now. – Pratyush Sarkar Nov 30 '13 at 16:08

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