First Order Differential Equation

  • Thread starter aolse9
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  • #1
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Homework Statement


Solve the following first order differential equation for x,

[tex]\frac{dy}{dx}[/tex] = 3xy + xy2

Homework Equations


Methods: Separation of Variables, Define an Integrating Factor



The Attempt at a Solution


I have been staring at this question for a while now, hoping that somehow the numbers would start looking nicer for me to solve, but alas this has not happened. I am totally stumped here, any help to steer me in the right direction would be greatly appreciated.
 

Answers and Replies

  • #2
525
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Start by seperating the left and right hand side, such that on one side you only have a y-dependence and on the other only an x-dependence. In other words, you want something like this:

[tex] f(y)dy = g(x) dx[/tex]

Written this way it's possible to integrate both sides (do you see why?). You can post your answer or attempt here.
 
  • #3
HallsofIvy
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That's separable: dy/dx= x(3y+ y2).
 
  • #4
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Thanks for the quick reply, so,

[tex]\frac{dy}{dx}[/tex] = x(3y + y2)

[tex]\frac{1}{3y + y^2}[/tex]dy = x dx

However, integrating the left hand side is beyond the scope of the subject I am taking at the moment. That's why they have asked to solve for x, I think. Is it still possible to solve for x using this method?
 
  • #5
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Any ideas?
 
  • #6
525
7
Are you sure about that? I don't really see an easier approach then to solve these integrals.

The integral on the left hand side is probably not 'out of your scope'. The trick is to seperate the fraction. It goes like this:

You have:
[tex]\frac{1}{3y+y^2} = \frac{1}{y(3+y)}[/tex]
Now make the following 'guess':
[tex]\frac{1}{y(3+y)} = \frac{a}{y}+\frac{b}{3+y}[/tex]
That is, we have split the fraction into two easier fractions. We just need to determine a and b -- I'll leave this up to you.

Answer:
[tex]\frac{1}{3y+y^2} = \frac{1/3}{y}-\frac{1/3}{3+y}[/tex]
You should be able to solve the integrals using this form.
 
  • #7
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Thanks for the quick reply, so,

[tex]\frac{dy}{dx}[/tex] = x(3y + y2)

[tex]\frac{1}{3y + y^2}[/tex]dy = x dx

However, integrating the left hand side is beyond the scope of the subject I am taking at the moment. That's why they have asked to solve for x, I think. Is it still possible to solve for x using this method?

If all you need to do is solve for x without solving the differential equation, you can do that in one step, starting with the 2nd equation above.
 
  • #8
23
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Ah OK, so you use the partial fraction method then integrate each term. Cheers xepma, I should be able to do it now. Also thanks for the input everyone.
 

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