Obtaining General Solution of ODE

Munir M
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Homework Statement


So they want me to obtain the general solution for this ODE.
Screen Shot 2016-11-01 at 12.00.43.png


Homework Equations


I have managed to turn it into d^2y/dx^2=(y/x)^2.

The Attempt at a Solution


My question is, can I simply make d^2y/dx^2 into (dy/dx)^2, cancel the power of 2 from both sides of the equation and then integrate it normally?

If not, why?
 
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I believe you can however, just make sure that you re-write it as dy/dx = +/- y/xAlso note the distinction that d^2y/dx^2 = d/dx(dy/dx) i.e. differentiating dy/dx wrt x and (dy/dx)*(dy/dx) = (dy/dx)^2.
 
rock.freak667 said:
I believe you can however, just make sure that you re-write it as dy/dx = +/- y/x
In other words, solve dy/dx = y/x as well as dy/dx = -y/x.
 
Mark44 said:
In other words, solve dy/dx = y/x as well as dy/dx = -y/x.

Thanks guys!
 
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...
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