Qualitative Differential equations

dismo
Messages
4
Reaction score
0
Problem:
Let x(t), y(t) e a solution of

dx/dt=y+x^2
dydt=x+y^2

with x(t0) NOT = y(t0)

Show that x(t) NOT = y(t) for all t

Attempt:

I feel like the easiest way to show this would be to show that x=y is an orbit of the system and then simply use the fact that orbits may not cross due to the uniqueness of IVP's at every point in the solution space?

So if I set x=y

dx/dt = y+y^2
dy/dt = y+y^2

This implies
dy/dx = 1

So y=x+c for all t.

Is this a reasonable solution?
Is there anything that needs clearing up?

Thanks guys.
 
Physics news on Phys.org
you started by assuming that y= x and showed that y= x+ c. How does that show that y= x is an orbit of the system? It might be better to determine a differential equation for y- x.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top