Solving Bernoulli ODE | Step-by-Step Guide | Example with n=-2

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I am working on the following Bernoulli ODE: 3xy^2 y' = 3x^4 + y^3. I come up with n = -2, so v = y^3 and y' = (1/3)v^(-2/3) v'. My integrating factor was x^-1. I end up with y^3 = X^2 + Cx yet the book has the same thing except X^4 instead of X^2. That makes me think I'm going wrong with multiplying it through but I don't see where.
 
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hi cue928! :smile:

(try using the X2 icon just above the Reply box :wink:)
cue928 said:
My integrating factor was x^-1. I end up with y^3 = X^2 + Cx yet the book has the same thing except X^4 instead of X^2.

it should have worked :confused:

show us how you got the x2
 
It was right, I just made a stupid mistake that led me to think it was something bigger than what it really was.
 
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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