Integrating factor

  • #1

Homework Statement



use an integrating factor to solve
[tex]
\frac{ \partial u}{ \partial x} = -2 + \frac{u}{2x}
[/tex]


The Attempt at a Solution


let P(x) = [tex] \frac {1}{2x} [/tex]

M(x) = [tex] e^(\int(\frac {1}{2x}dx)) [/tex]

= [tex]\sqrt{x}[/tex]

so u = [tex]
\frac{1}{ \sqrt{x}} \int-2\sqrt{x} dx = \frac{1}{ \sqrt{x}}(\frac{2}{3} x^(3/2))
[/tex]

=-4x + k


anyone got any ideas if im doing this right?
thanks
 

Answers and Replies

  • #2
fzero
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You have a numerical integration factor in the wrong place

so u = [tex]
\frac{1}{ \sqrt{x}} \int-2\sqrt{x} dx = \frac{1}{ \sqrt{x}}(\frac{2}{3} x^{(3/2)} + \mathbf{k})
[/tex]

[tex] =-4x + \mathbf{\frac{k}{\sqrt{x}}}[/tex]
 
  • #3
deadly thanks a mill
 
  • #4
sorry should it be [tex]
=\frac {-4x}{3} + \mathbf{\frac{k}{\sqrt{x}}}
[/tex]
 
  • #5
fzero
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Sorry, I looked at the -4x in your solution, which is correct, so I missed an earlier mistake. You should have had

[tex] P(x) = -\frac{1}{2x}.[/tex]

This changes quite a bit, so you should go back through the algebra.
 
  • #6
i dont understand why [tex]
P(x) = -\frac{1}{2x}.
[/tex] should it not be term infront of u?
 
  • #7
fzero
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i dont understand why [tex]
P(x) = -\frac{1}{2x}.
[/tex] should it not be term infront of u?

[tex]P(x)[/tex] is defined with that term on the same side as the derivative:

[tex] u'(x) + P(x) u(x) = q(x).[/tex]

Remember that the integrating factor works because

[tex](u'(x) + P(x) u(x) ) \exp \left(\int P(x) dx\right) = \frac{d}{dx} \left[ u(x) \exp \left(\int P(x) dx \right) \right]. [/tex]
 
  • #8
ahh yes i didnt bring it across, thanks
 
  • #9
i think this is right?
let P(x) = [tex] \frac {-1}{2x} [/tex]

M(x) = [tex] e^(\int(\frac {-1}{2x}dx)) [/tex]

= [tex]\frac{1}{\sqrt{x}}[/tex]

so u = [tex]
\sqrt{x} \int \frac {-2}{\sqrt{x}} dx = \sqrt{x}(-4\sqrt{x} + k)
[/tex]

=-4x + [tex]\sqrt{x}[/tex] k
 
  • #10
fzero
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The easiest way to verify a solution to a differential equation is to explicitly check that it actually satisfies the differential equation.
 
  • #11
i cant figure out where i went wrong!
 
  • #12
dextercioby
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You didn't go wrong. I found the same solution

[tex] u(x)= -4x + C\sqrt{x} [/tex]
 
  • #13
oh, ok cool. thanks a mill for your help...you have great patience
 

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