Existence and Uniqueness

Question

Please I can't seem to figure this problem out,

Determine the interval of existence of the solution $y'=x^2 + y^2$, $y(1)=3$.

Thanks

I need to solve it using the idea of Uniqueness and Existence Theorem. Thanks in advance — Soledolu Oluwatayo, Feb 04 '17 at 21:10
What exactly are you supposed to compute? This equation is not simple to solve, see https://math.stackexchange.com/q/446926/115115, and will have a pole before $x-1+\arctan(3)\le\frac\pi2$. Do you need the existence of that pole, or better approximations of its positions or ...? See also https://math.stackexchange.com/q/1353727/115115 for ideas on estimates. — Lutz Lehmann, Feb 04 '17 at 21:54

score 0 · Answer 1 · answered Feb 04 '17 at 20:40

0

Just find the interval where $x^2+y^2$ and $\frac{\partial}{\partial y}(x^2+y^2)=2y$ is continuous.

answered Feb 04 '17 at 20:40

bellcircle

3,069

Are you saying that the solution is global? – John B Feb 04 '17 at 21:40
This problem needs to be solved using the definition of the Uniqueness and Existence Theorem. Any ideas will be appreciated thanks – Soledolu Oluwatayo Feb 08 '17 at 01:19

score 0 · Answer 2 · answered Dec 26 '17 at 11:10

For $x\ge 1$, $$ y'\ge 1+y^2\implies\arctan(y(x))-\arctan(y(1))\ge x-1\implies y(x)\ge \tan(x-1+\arctan(3)) $$ the last is maximally valid to the first pole of the tangent after $x=1$, that is to $x=1+\frac\pi2-\arctan(3)=1.32175...$. The solution $y$ has to reach infinity before that point.

As $y'\ge 0$ one also gets $y\ge 3$ for $x\ge 1$ so that $$ y'\le(\frac19x^2+1)y^2\implies \frac1{y(1)}-\frac1{y(x)}\le \frac1{27}(x^3-1)+(x-1) \\ y(x)\le\frac{27}{37-27x-x^3} $$ which means $y$ remains finite to the first root of the denominator at $\approx 1.290728579...$

Existence and Uniqueness

2 Answers2