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November 21st, 2011, 01:29 PM   #11
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Re: differential equation

I know I'm asking for much, but this is the last one I'm asking you guys to help me with
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 November 21st, 2011, 01:31 PM #12 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: differential equation For the one you uploaded the image of your working: We are given: $\frac{dy}{dx}=e^{x-y}=\frac{e^x}{e^y}$ Separate variables: $e^y\,dy=e^x\,dx$ $\int e^y\,dy=\int e^x\,dx$ $e^y=e^x+C$ Convert from exponential to logarithmic form: $y=\ln$$e^x+C$$$
November 21st, 2011, 04:52 PM   #13
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From: St. Augustine, FL., U.S.A.'s oldest city

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Math Focus: Calculus/ODEs
Re: differential equation

Quote:
 Originally Posted by BonTrust I know I'm asking for much, but this is the last one I'm asking you guys to help me with
We are given:

$$$1+y^2$$$$e^{2x}\,dx-e^y\,dy$$-$$1+y$$dy=0$

Divide through by $$$1+y^2$$$:

$e^{2x}\,dx-e^y\,dy-\frac{1+y}{1+y^2}dy=0$

Separate variables:

$e^{2x}\,dx=e^y\,dy+\frac{1+y}{1+y^2}dy$

$\frac{1}{2}e^{2x}\,2dx=e^y\,dy+\frac{1}{y^2+1}\,dy +\frac{1}{2}\cdot\frac{2y}{y^2+1}\,dy$

$\frac{1}{2}\int e^{2x}\,2dx=\int e^y\,dy+\int\frac{1}{y^2+1}\,dy+\frac{1}{2}\int\fr ac{2y}{y^2+1}\,dy$

The solution can then be given implicitly by:

$\frac{1}{2}e^{2x}+C=e^y+\tan^{\small{-1}}$$y$$+\frac{1}{2}\ln$$y^2+1$$$

November 21st, 2011, 09:46 PM   #14
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Re: differential equation

Quote:
Originally Posted by MarkFL
Quote:
 Originally Posted by BonTrust I know I'm asking for much, but this is the last one I'm asking you guys to help me with
We are given:

$$$1+y^2$$$$e^{2x}\,dx-e^y\,dy$$-$$1+y$$dy=0$

Divide through by $$$1+y^2$$$:

$e^{2x}\,dx-e^y\,dy-\frac{1+y}{1+y^2}dy=0$

Separate variables:

$e^{2x}\,dx=e^y\,dy+\frac{1+y}{1+y^2}dy$

$\frac{1}{2}e^{2x}\,2dx=e^y\,dy+\frac{1}{y^2+1}\,dy +\frac{1}{2}\cdot\frac{2y}{y^2+1}\,dy$

$\frac{1}{2}\int e^{2x}\,2dx=\int e^y\,dy+\int\frac{1}{y^2+1}\,dy+\frac{1}{2}\int\fr ac{2y}{y^2+1}\,dy$

The solution can then be given implicitly by:

$\frac{1}{2}e^{2x}+C=e^y+\tan^{\small{-1}}$$y$$+\frac{1}{2}\ln$$y^2+1$$$
I thought the pattern was to separate the y at the end, to get something like y = x...+c, not another equation:/ I'm confused.

 November 21st, 2011, 09:55 PM #15 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: differential equation Sometimes, you get an implicit relationship between the dependent and independent variables, rather than an explicit one. With this one, we could solve for x, to get x as a function of y...
 November 22nd, 2011, 02:43 AM #16 Global Moderator   Joined: Dec 2006 Posts: 20,833 Thanks: 2161 There was nothing in the original equation to suggest that y was the dependent variable and x the independent variable.

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