Solving a Diff. Eq. Problem: y = (2x^-1 + Cx^4)^-1/2

  • Thread starter Thread starter cue928
  • Start date Start date
cue928
Messages
129
Reaction score
0
(x^2)y' + 2xy = 5y^3
y' + (2/x)y = 5(x^-2)(y^3) v=y^-2, y=v^(-1/2), y' = -(1/2)v^(-3/2)
-(1/2)v^(-3/2)v' + (2/x)v^(-1/2) = 5x^-2(v^-3/2)
v' - (4/x)v = -10x^-2 Integrating factor: x^-4
(skipping a few mechanical steps)
vx^-4 = 2x^-5 + C
v = 2x^-1 + Cx^4
I come up with y = (2x^-1 + Cx^4)^-1/2

Problem is the book shows y^2 = x/(2+Cx^5)? What am I missing?
 
Physics news on Phys.org
y = (2x^-1 + Cx^4)^-1/2 and y^2 = x/(2+Cx^5) are equivalent. Multiply the first expression by \sqrt{x/x}.
 
OMG, I am an idiot. I should have seen that.
 
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