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November 21st, 2011, 09:22 AM   #1
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differential equation

can anyone help me with this?:/
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 November 21st, 2011, 09:50 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: differential equation $\frac{dy}{dx}= sin(y)$ $\frac{dy}{sin(y)}= dx$ $\csc(y)\,dy= dx$ Integrate both sides.
 November 21st, 2011, 09:56 AM #3 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 Don't neglect the trivial solution y = k?, where k is an integer.
 November 21st, 2011, 10:20 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,020 Thanks: 2255 For y ? k?, let u = tan(y/2), then du/dx = (1/2)sec˛(y/2)dy/dx, so that dy/dx = 2(du/dx)/(1 + u˛). Hence the equation becomes 2du/dx/(1 + u˛) = 2u/(1 + u˛), i.e., du/dx = u, which has solution u = Ce^x. Therefore y = 2arctan(Ce^x) + 2k? or y = k? (as noted above), where C is a constant and k is an integer.
November 21st, 2011, 12:10 PM   #5
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Re: differential equation

Quote:
 Originally Posted by MarkFL Don't neglect the trivial solution y = k?, where k is an integer.
y' = sin(y)
Cancelling y from both sides gives:
' = sin

Something seems fishy about my version of the trivial solution

November 21st, 2011, 12:32 PM   #6
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Re: differential equation

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November 21st, 2011, 12:35 PM   #7
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Re: differential equation

Quote:
Maybe a substitution can be useful ...

 November 21st, 2011, 12:49 PM #8 Newbie   Joined: Oct 2011 Posts: 17 Thanks: 0 Re: differential equation Well, that's the point... I'm very bad at math :/ I know to code web sites with html, css, jQuery, php, create programs in Java, knowing design patterns, uml and sql, but math is just too hard for me:/
 November 21st, 2011, 01:07 PM #9 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 We are given: $$$1+x^2$$\frac{dy}{dx}-2xy=$$1+x^2$$^2$ If we write it like this: $\frac{$$1+x^2$$\frac{dy}{dx}-2xy}{$$1+x^2$$^2}=1$ Observing we have the quotient rule on the left, we may then write: $\frac{d}{dx}$$\frac{y}{1+x^2}$$=1$ $\int\,d$$\frac{y}{1+x^2}$$=\int\,dx$ $\frac{y}{1+x^2}=x+C$ $y=$$1+x^2$$$$x+C$$$
November 21st, 2011, 01:24 PM   #10
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Re: differential equation

Thanks it really helps to read some steps on how to do it
I have a question? Is this solution to other equation right? I was trying to do something by myself
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