Question

The initial value problem dy/dx = x^2+y^2, y(0) = 0 has a unique solution in some interval of the form -h<x<h. True or false ,why?

Answer 1

Answer on Question #66105 – Math – Calculus

Question

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 $-h < x < h$.

True or false, why?

Solution

**Existence and Uniqueness Theorem for First Order ODE’s [1, page 150]:**

Let $R$ be a rectangle and let $f(x, y)$ be continuous throughout $R$ and satisfy the Lipschitz Condition with respect to $y$ throughout $R$. Let $(x_0, y_0)$ be an interior point of $R$. Then there exists an interval containing $x_0$ on which there exists a unique function $y(x)$ satisfying $y' = f(x, y)$ and $y(x_0) = y_0$.

We have

$$f(x, y) = x^2 + y^2; \quad \frac{df}{dy} = 2y,$$

hence $f(x, y)$ has a continuous derivative with respect to $y$, therefore $f(x, y)$ satisfies the Lipschitz Condition with respect to $y$ throughout rectangle $R$ [2, Proposition 1].

Since $f(x, y)$ satisfies the conditions of Existence and Uniqueness Theorem, 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 $-h < x < h$.

**Answer:** true.

References:

[1] Differential Equations I, MATB44H3F, Version September 15, 2011-1949.

[2] Lipschitz condition and differentiability. Retrieved from http://planetmath.org/lipschitzconditionanddifferentiability

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