Question

The initial value problem
x y , )0(y 0
dx
dy 2 2
= + =
has a unique solution in some interval of the form − h < x < h .

Answer 1

Answer on Question 66455 - Math - Differential Equations

The initial value problem

$\frac{dy}{dx}=x^{2}+y^{2},\quad y(0)=0$

has a unique solution in some interval of the form $-h<x<h$.

Solution The well-known existence and uniqueness theorem states: if the right-hand $f(x,y)$ of the differential equation $y^{\prime}=f(x,y)$ is a continuously differentiable function on some domain $\Omega$, then for any initial data $(x_{0},y_{0})\in\Omega$ we can find a neighbourhood $[x_{0}-h,x_{0}+h]$ in which there exists an unique solution $y=y(x)$ of the equation such that $y(x_{0})=y_{0}$.

In our case we have $f(x,y)=x^{2}+y^{2}$, $x_{0}=0$, $y_{0}=0$. The function $f$ is continuously differentiable in $\mathbb{R}^{2}$, since its partial derivatives

$\frac{\partial f}{\partial x}(x,y)=2x,\qquad\frac{\partial f}{\partial y}(x,y)=2y$

are continuous functions in $\mathbb{R}^{2}$. Therefore the initial problem

$\frac{dy}{dx}=x^{2}+y^{2},\quad y(0)=0$ (1)

has a unique solution in an interval $-h<x<h$ with some positive $h$. In addition, we can find the concrete number $h$ with such property. For instance, we consider the rectangle

$R=\{(x,y)\in\mathbb{R}^{2}\colon|x|\leq a,\quad|y|\leq b\}$

and set

$M=\max_{(x,y)\in R}|f(x,y)|.$

It is known that then a solution of (1) exists in the interval $-h<x<h$, where

$h=\min\left\{a,\frac{b}{M}\right\}$

We have

$M=\max_{(x,y)\in R}(x^{2}+y^{2})=a^{2}+b^{2}\quad\Rightarrow\quad h=\min\left\{a,\frac{b}{a^{2}+b^{2}}\right\}$

Set $a=1$ and $b=1$. Then $h=\min\{1,0.5\}=0.5$. Hence, we can state that the solution $y=y(x)$ of (1) exists in the interval $-0.5<x<0.5$.

Answer: The statement is true. There exists an unique solution $y=y(x)$ of (1) in the interval $-0.5<x<0.5$.