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September 30th, 2012, 10:40 AM   #1
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Exact differential equation

Hey guys, I've been trying to solve a differential equation, I checked 20 times but it's not the right answer.. It's

dy/dx=(3*x^3+y)/(-x-4*y^3) wich equals (3*x^3+y)dx+(x+4*y^3)dy=0

So My=1 and Nx=1, so the equation is exact

Then I do the integer of 3*x^3+y, which gives 3/4*x^4+x*y+(a function of y)

I do all the steps and I find the function of y, which is y^4

so my answer is 3/4*x^4+x*y+y^4 but it says it's not the right answer. Could anyone help me please?
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September 30th, 2012, 10:57 AM   #2
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Re: Exact differential equation

You have correctly expressed the ODE in differential form:

You have correctly determined the equation is exact.

So, we set:

Now differentiate with respect to y:

Hence, the solution is given implicitly by:

It appears the only thing you have left out is the constant.
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September 30th, 2012, 11:14 AM   #3
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Re: Exact differential equation

That's what I thought but I keep entering this answer and it says that it's wrong... maybe there's a problem on the site..
Thanks a lot!
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October 1st, 2012, 07:02 PM   #4
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Re: Exact differential equation

The way I would have done this problem is this: to say that is an exact equation means that there exist a function F(x,y) such that .

So we must have and differentiating with respect to x treats y as a constant, integrating, treating y as a constant, where, because we are treating y as a constant, the "constant of integration" may be a function of y, g(y). Differentiating that with respect to y, . That is, we must have so that where, because g is a function of y only, C really is a constant.

That is . Because the original differential equation was "dF= 0", F is a constant: and we can combine the two constant to get .
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