Overview

• Theorem on existence of solution to IVP

• Examples

Theorem. If $f$ continuous on the open rectangle $R:a<x<b,c<y<d$, and $(x_0,y_0)\in R$, then the IVP $y^{\text{'}}=f(x,y)$, $y\left(x_0\right)=x_0$ has at least one solution on some open interval of $(a,b)$ that contains $x_0$

Theorem. If $f$ and $f_y=\partial f/\partial y$ continuous on the open rectangle $R:a<x<b,c<y<d$, and $(x_0,y_0)\in R$, then the IVP $y^{\text{'}}=f(x,y)$, $y\left(x_0\right)=x_0$ has an unique solution on some open interval of $(a,b)$ that contains $x_0$

${\displaystyle y^{\text{'}}=\frac{x^2-y^2}{1+x^2+y^2},y\left(x_0\right)=y_0}$

${\displaystyle y^{\text{'}}=\frac{x^2-y^2}{x^2+y^2},y\left(x_0\right)=y_0}$

${\displaystyle y^{\text{'}}=\frac{x+y}{x-y},y\left(x_0\right)=y_0}$

${\displaystyle y^{\text{'}}=2x{y}^2,y\left(x_0\right)=y_0}$

${\displaystyle y^{\text{'}}=\frac{10}{3}x{y}^{2/5},y\left(x_0\right)=y_0}$

${\displaystyle y^{\text{'}}=\frac{10}{3}x{y}^{2/5},y\left(0\right)=0}$

${\displaystyle \begin{array}{rcl}{y}^{\text{'}}=\frac{10}{3}x{y}^{2/5},y\left(0\right)=-1& & \end{array}}$

General solution: $y^{3/5}=x^2+c\Rightarrow y={(x^2+c)}^{5/3}$.

Initial condition: $y\left(0\right)=c^{5/3}=-1$ has no solution in $ℝ$.

However in $ℂ$, $c^{5/3}=e^{i\pi }\Rightarrow c=e^{3i\pi /5}$