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 June 30th, 2013, 08:26 AM #1 Newbie   Joined: Sep 2012 Posts: 7 Thanks: 1 ODEs that are not quite simple for me. Hello, I am not really great at arithmetic and just an average joe but this equations are killing me. I tried doing everything I know but nothing makes sense. Can anybody help me to find C in the following ODEs? 1. dy/dx = y/x 2. dy/dx = y-x^2 Thank you for your response.  June 30th, 2013, 09:01 AM   #2
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Re: ODEs that are not quite simple for me.

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 Originally Posted by Execross02 Can anybody help me to find C in the following ODEs? 1. dy/dx = y/x 2. dy/dx = y-x^2
Where do you see C in those ODEs ?
What do you mean ? Can you explain what C is ? June 30th, 2013, 09:04 AM #3 Newbie   Joined: Sep 2012 Posts: 7 Thanks: 1 Re: ODEs that are not quite simple for me. we have to integrate both sides right? so in the first problem for example dy/dx = y/x dy/y=dx/x then to integrate: ln y = ln x + C the second problem is problematic because I get y= xy - x^3(1/3) + C June 30th, 2013, 09:15 AM #4 Math Team   Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: ODEs that are not quite simple for me. If no initial values are given, C is not expressible in terms of any number on R. However, one can do this : log y(x) = log(x) + C implies C = log y(1). And in y(x) = x * y(x) - x^3/3 + C, C = y(0). June 30th, 2013, 09:44 AM #5 Newbie   Joined: Sep 2012 Posts: 7 Thanks: 1 Re: ODEs that are not quite simple for me. You are supposed to let x = n. so any whone number can replace x. then we could solve for y. a friend of mine told me you are supposed to use integrating factors. on the 2nd problem. I got y=1+c. June 30th, 2013, 09:53 AM   #6
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Re: ODEs that are not quite simple for me.

Quote:
 Originally Posted by Execross02 You are supposed to let x = n. so any whone number can replace x. then we could solve for y.
Solve for y or solve for C? If the later, that's what I did, set x = 1.

Quote:
 Originally Posted by Execross02 a friend of mine told me you are supposed to use integrating factors. on the 2nd problem.I got y=1+c.
What? June 30th, 2013, 10:01 AM   #7
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Re: ODEs that are not quite simple for me.

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Originally Posted by mathbalarka
Quote:
 Originally Posted by Execross02 You are supposed to let x = n. so any whone number can replace x. then we could solve for y.
Solve for y or solve for C? If the later, that's what I did, set x = 1.

isn't this what you gonna do? like 2=1+c then c=2-1, C=1.

Quote:
 Originally Posted by Execross02 a friend of mine told me you are supposed to use integrating factors. on the 2nd problem.I got y=1+c.
What?
this: June 30th, 2013, 10:04 AM   #8
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Re: ODEs that are not quite simple for me.

Quote:
 Originally Posted by Execross02
Sorry, my internet speed is too slow to view a youtube. June 30th, 2013, 10:08 AM #9 Newbie   Joined: Sep 2012 Posts: 7 Thanks: 1 Re: ODEs that are not quite simple for me. oh... it is when y' + p(x)=q(x) then you multiply all sides by e^\$p(x)dx July 1st, 2013, 08:15 AM #10 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: ODEs that are not quite simple for me. dy/dx= y- x^2 can be written as dy/dx- y= x^2, a linear equation. An "integrating factor" for a this equation is a function, u(x), such that d(uy)/dx= u dy/dx+ uy. Using the product rule, u dy/dx+ u' y= udy/dx- uy. That tells us u'= -u which has solution u(x)= e^(-x). That is, the equation is of the form e^x dy/dx- e^x y= d(e^x y)/dx= x^2e^x. e^x y= . That last integral can be done by "integration by parts". Tags odes, simple Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zibb3r Calculus 1 September 21st, 2013 01:35 PM singapore Calculus 2 March 19th, 2012 06:48 PM FreaKariDunk Calculus 12 February 22nd, 2012 08:59 PM liakos Applied Math 0 January 1st, 2011 09:52 AM acnash Calculus 2 March 25th, 2010 04:19 PM

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