find_the_fun
- 147
- 0
Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point [math](x_0, y_0)[/math] in the region.
[math]x \frac{dy}{dx} = y[/math]
What does an xy-plane have to do with anything? I looked up the definition of unique solutions and here it is
Let R be a rectangular region in the xy-planed defined by a <=x<=b, c<=y<=d that contains the point [math](x_0, y_0)[/math] in its interior. If f(x,y) and [math]\frac{\partial{d} f}{\partial{d} y}[/math] are continuous on R then there exists some interval [math]I_0: (x_0-h, x_0+h), h>0[/math] contained in [a/b] and a unique function y(x) defined on [math]I_0[/math] that is a solution of the initial value problem.
That's a bit difficult to digest. How do I proceed?
[math]x \frac{dy}{dx} = y[/math]
What does an xy-plane have to do with anything? I looked up the definition of unique solutions and here it is
Let R be a rectangular region in the xy-planed defined by a <=x<=b, c<=y<=d that contains the point [math](x_0, y_0)[/math] in its interior. If f(x,y) and [math]\frac{\partial{d} f}{\partial{d} y}[/math] are continuous on R then there exists some interval [math]I_0: (x_0-h, x_0+h), h>0[/math] contained in [a/b] and a unique function y(x) defined on [math]I_0[/math] that is a solution of the initial value problem.
That's a bit difficult to digest. How do I proceed?