Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $-2 < t <2$
share|improve this question
2  
Please read about accepting answers here and here. –  Git Gud Nov 30 '13 at 13:17

1 Answer 1

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 $c$. $$ h(x) = \begin{cases} f(x) & -2 < x \leq 0 \\ ct^2 & 0 \leq x < 2 \end{cases} $$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.