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I am doing this problem. For part a, I think I can just let $y(t)=at^2+bt+c$, but the problem is, how can I make sure what is the max precision of $y(t)$? I am not sure if it is of degree 2 or 3 or 4, but since I am only asked to find $y(0)=s$, I suppose it doesn't matter?

And for part b, can anyone please help? Thanks a lot!

J.doe
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1 Answers1

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Your solution is wrong. Just by inserting it you will find that a quadratic polynomial is no solution.

"Blowing up in finite time" means that the solution runs into a singularity, most often a pole. A quadratic polynomial does not have a pole.

You can find the exact solution by separation of variables and partial fraction decomposition or by treating the ODE as a Bernoulli equation.

Using $y=u^a$, $y'=au^{a-1}u'$ you get $$ au^{a-1}u'=\frac12 (u^a+u^{3a})\iff au'=\frac12 (u+u^{2a+1}) $$ which simplifies for $a=-\frac12$ to $$ u'=-u-1\iff u=-1+Ce^{-t}\implies y(t)=\frac1{\sqrt{Ce^{-t}-1}} $$ so that $C-1=s^{-2}$ and consequently the singularity appears at $t=\ln(C)=\ln(1+s^{-2})$.

Lutz Lehmann
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  • Can you talk more about the blowing up part? I am not sure that I get it. Thanks! – J.doe Mar 22 '16 at 18:54
  • The solution ceases to exist for this concrete example since the quantity below the root sign changes at $t=\ln(1+s^{-1})$ to the negative sign. Since it crosses zero, the value of $y(t)$ approaches infinity at this point. There is nothing more to it. – Lutz Lehmann Mar 22 '16 at 19:02
  • I got it! Thank you very much! – J.doe Mar 22 '16 at 19:04