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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.

Quote:
 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.

Quote:
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= $\int x^2e^x dx$. That last integral can be done by "integration by parts".

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