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 February 11th, 2012, 02:36 AM #1 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 Differentiation After differentation dy/dx = x+y will become, to find the exact solution y = e^x - x - 1 I need to do the same to dy/dx = y^3 (1-x^2) Can anyone help?
 February 11th, 2012, 03:41 AM #2 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: Differentiation The second ODE is separable. $\frac{dy}{dx}=y^3(1-x^2)$ $\Leftrightarrow \frac{dy}{y^3}=(1-x^2)dx$ $\Leftrightarrow \int \frac{dy}{y^3}=\int (1-x^2)dx$ Finish it ...
 February 11th, 2012, 08:33 AM #3 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 Re: Differentiation Would it be -1/y=x-x^3/3
 February 11th, 2012, 10:26 AM #4 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: Differentiation For the first one you have $y' = y + x$ Put the equation like this $y' - y = x$ Multiply by $e^{-x}$ $y'e^{-x} - ye^{-x} = xe^{-x}$ $d(y e^{-x})= xe^{-x}dx$ $y e^{-x}= -xe^{-x}-e^{-x}+C$ $y= -x-1+Ce^x$ Notice we don't exclude the -1 as an arbitrary constant since it is part of the particular solution $y_p=-(x+1)$ to the equation. $Ce^x$ is the solution to the homogeneous equation.
 February 12th, 2012, 04:25 AM #5 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 Re: Differentiation Thanks for that but I need a bit more help with the second
 February 12th, 2012, 09:58 AM #6 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: Differentiation For the second you separate variables, putting $\frac{dy}{dx}= y^3(1-x^2)$ $\frac{dy}{y^3}= (1-x^2)dx$ $-y^{-2}= (x-\frac{x^3}{3})+C$ $y^{-2}= (\frac{x^3}{3}-x)+C_1$ $y= \frac{1}{\sqrt{(\frac{x^3}{3}-x)+C_1}}$
 February 12th, 2012, 01:15 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 Or y = 0, which is the only real function that satisfies the equation for all real values of x.

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