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 November 11th, 2011, 12:24 AM #1 Newbie   Joined: Nov 2011 Posts: 4 Thanks: 0 Differential equation simplification Hi. Just need some help understanding how my tutor simplified a differential equation and what laws he used to get his answer. The question is: e^(y/x)y'=2(e^(y/x)-1)+(y/x)e^(y/x) - use the substitution y=xv(x) to solve this. I can post my whole working if you need to see it but basically both mine and the tutors answers simplify to this point ln|e^v-1|=ln|x^2|+c Now i would simplify this to e^v-1=x^2+e^c but I am clearly doing something wrong because my tutor gets ln|e^v-1|=x^2*e^c which then somehow simplifies to cx^2 some basic law seems to be hovering just out of reach. Would appreciate an explanation. Thanks!
 November 11th, 2011, 03:19 AM #2 Newbie   Joined: Oct 2011 Posts: 26 Thanks: 0 Re: Differential equation simplification $\ln|e^v-1|=\ln|x^2|+c$ $e^v-1= e^{\ln |x^2|+c} = e^{\ln|x^2|}e^c = x^2\cdot e^c = c_1x^2$ where $c_1=e^c$ is a constant
 November 12th, 2011, 08:25 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,105 Thanks: 2324 If y is defined for all real x, $y\,=\,\begin{cases} x\ln(1\,+\,\,\text{A}x^{\small2}) &\;\;x\,\le\,0 \\ x\ln(1\,+\,\,\text{B}x^{\small2})=&\;\;x\,\,=>\,\,0\end{cases}=$  where the constants A and B are non-negative. I'll leave you to specify the domain if the constants are not both non-negative.

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