Question

determine a region of the xy-plane for which the given DE would have a unique solution whose graph passes through a point (x0,y0) in the region

(y-x)y'=y+x

Answer 1

Answer on Question # 70254 - Math - Differential Equations

Question

Determine a region of the xy-plane for which the given DE would have a unique solution whose graph passes through a point $(x_0, y_0)$ in the region $(y - x)y' = y + x$

Solution

We have the initial value problem

$$y' = \frac{y + x}{y - x} \qquad y(x_0) = y_0.$$

To determine a region of the xy-plane for which the given DE would have a unique solution whose graph passes through a point $(x_0, y_0)$ in the region, we use the Fundamental Theorem of Existence and Uniqueness for a first order differential equation

$$y' = f(x, y), y(x_0) = y_0.$$

If $f(x, y)$ and $\frac{\partial f}{\partial y}$ are continuous on a rectangular region defined by $a < x < b$, $c < y < d$ that contains the point $(x_0, y_0)$, then there exist an interval centered at $x_0$ and a unique function $y(x)$ defined on the interval that satisfies the Initial Value Problem.

For this problem

$$f(x, y) = \frac{y + x}{y - x}$$

then

1. $f(x, y) = \frac{y + x}{y - x}$ is continuous when $y \neq x$ or otherwise $y > x$ or $y < x$.

2. $\frac{\partial f}{\partial y} = \begin{pmatrix} y + x \\ y - x \end{pmatrix}_y = \frac{(y + x)_y \cdot (y - x) - (y + x) \cdot (y - x)_y}{(y - x)^2} = \frac{1 \cdot (y - x) - (y + x) \cdot 1}{(y - x)^2} = \frac{-2x}{(y - x)^2}$

also is continuous when $y > x$ or $y < x$.

Therefore, by the Fundamental Theorem of Existence and Uniqueness, in the region of $y > x$ or $y < x$ the given DE would have a unique solution whose graph passes through a point $(x_0, y_0)$ in the region.

**Answer**: a region of the xy-plane for which the DE $(y - x)y' = y + x$ would have a unique solution whose graph passes through a point $(x_0, y_0)$ in the region, is $y > x$ or $y < x$.

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