Help with nonlinear 1st order ODE

[X89codered89X]

So I'm supposed to prove that
${x}^{.}(t) = x^{2}+ t^{2}$ with $x(0) = 0$ blows up before $t = 1$.

I'm not sure what method to use to solve I've tried setting up an integral such as $\int^{x(t)}_{x(0)} \frac{dx}{x^{2}+t^{2}} = \int^{t}_{0} dt$ but I didn't think I could do this since 't' is varying over time on the Left Hand side and I'm integrating with respect to x.

The only other clue I have is to use a comparison ODE, which was mentioned in class, in which would use a function, say... $g^{.}(t) =< x^{.}(t)$ which was easier to work with. If I were able to prove that the lesser function exploded before $t= 1$, then logically the greater one explodes. The thing is I don't know what function I would even chose to set this up. Any ideas?

..and Thank youuuuu.

 

[JJacquelin]

Hello !

The ODE is on the Riccati kind. First let x(t) = -y'/y
leading to y''+t²y=0
which is on the Bessel kind
y(t) = sqrt(t)*(c1*J(k,X)+c2*J(-k,X) )
J(k,X) is the Bessel function of order k
k=1/4
X=t²/2
k=1/4

 

[X89codered89X]

Right OK! can I then substitute that back in somehow to show that it explodes at t -> 1? I don't necessarily need to get it solved. Just know how it behaves.

 

[JJacquelin]

Right OK! can I then substitute that back in somehow to show that it explodes at t -> 1? I don't necessarily need to get it solved. Just know how it behaves.

From my results it doesn't "explodes" close to t=1, but at t=2.003147 which is the first root of the BesselJ function of order -1/4.

 

[blue_raver22]

Hello, thank you for reaching out for help with your nonlinear first-order ODE. To prove that the solution blows up before t=1, you can use the method of finding a lower bound for the solution. This involves finding a function g(t) such that g'(t) < x'(t) for all t, and g(t) < x(t) for t=0.

One way to find this lower bound is by using the comparison ODE method you mentioned. You can choose a function g(t) that is simpler and easier to work with, such as g(t) = t. Then, you can compare the derivatives g'(t) and x'(t) to establish that g'(t) < x'(t) for all t, and g(0) = 0 < x(0) = 0.

Next, you can solve the comparison ODE g'(t) = g(t)^2 + t^2 to find the solution g(t) = tan(t^2/2). Then, you can show that this solution blows up at t=1, since g(1) = tan(1/2) is not defined.

Since g(t) is a lower bound for x(t), this means that x(t) must also blow up before t=1. This proves that the solution x(t) = tan(t^2/2) blows up before t=1 as well.

I hope this helps. Good luck with your proof!