Rearranging separable equations

  • Thread starter Thread starter thedude36
  • Start date Start date
  • Tags Tags
    Separable
thedude36
Messages
30
Reaction score
0
I'm having difficulty with the question in the pic provided. (http://i.imgur.com/Fg7CHoY.png). Basically the equation I am given is

dy/dx = (y-4x)/(x-y)​

and it walks through the steps needed to solve it, however, I am supposed to show how to arrive at each step given. I've gotten part a, where it asks to show that the above equation can be written as

dy/dx = (y/x - 4)/(1-y/x)​

and, introducing v=y/x, rewrite dy/dx in terms of v, x, and dv/dx. What I get for part b is

dy/dx = d(xv)/dx = x(dv/dx)​

however, on part c, it looks like I'm supposed to get

dy/dx = v+x(dv/dx)​

I'm not sure where the additional v term comes from. If anyone could help me out, i would really appreciate it!
 
Physics news on Phys.org
thedude36 said:
dy/dx = d(xv)/dx = x(dv/dx)
dy/dx = v+x(dv/dx)​
It's just the product rule. How do you differentiate f(x)g(x)?
 
I can't believe it was that simple - I had been staring at it for hours. Thank you!
 
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