Solving an Arrogant ODE: Can You Do It?

  • Thread starter Thread starter Altabeh
  • Start date Start date
  • Tags Tags
    Ode
Altabeh
Messages
657
Reaction score
0

Homework Statement



How can you solve an arrogant ODE of form {\frac {\left( {\frac {d}{dt}}\rho \left( t \right) \right) ^{2}}{\rho \left( t \right) }}=-3??

The Attempt at a Solution



I don't have any idea... maybe you do!

Thanks
AB
 
Physics news on Phys.org
From this you get (\rho '(t))^2 = -3\rho(t) from which you get two equations

\rho '(t) = +\sqrt{-3\rho(t)}

and

\rho '(t) = -\sqrt{-3\rho(t)}

Both are separable.
 
Mark44 said:
From this you get (\rho '(t))^2 = -3\rho(t) from which you get two equations

\rho '(t) = +\sqrt{-3\rho(t)}

and

\rho '(t) = -\sqrt{-3\rho(t)}

Both are separable.

Sorry. I'm a little bit confused right now, but how would this separation help us to get \rho(t)?

AB
 
If you divide both sides by sqrt(rho), you get:

rho'/sqrt(rho) = +/-sqrt(3) i

You can write this as:

d rho/sqrt(rho) = +/-sqrt(3) i dt

You can now integrate both sides.
 
a simple ODE with a complex answer :D
 
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