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June 30th, 2013, 08:26 AM   #1
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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.
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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 ?
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June 30th, 2013, 09:04 AM   #3
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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
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June 30th, 2013, 09:15 AM   #4
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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).
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June 30th, 2013, 09:44 AM   #5
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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.
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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?
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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:
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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.
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June 30th, 2013, 10:08 AM   #9
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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
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July 1st, 2013, 08:15 AM   #10
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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".
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