Solving an ODE

  • Thread starter cragar
  • Start date
  • #1
cragar
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Homework Statement


Ay'+Bxy=Cy
y=f(x)
A,B,C are real constants

The Attempt at a Solution


This kinda looks like a Bernoulli equation but not really.
I thought about using an integrating factor but there is function of x on the right side.
If I tried undetermined coefficients what would my guess function be.
 

Answers and Replies

  • #2
elvishatcher
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0
It seems like you could solve this by first manipulating to get [tex]y' + \frac{Bx-C}{A}y = 0[/tex] at which point you now have an ODE of the form [tex]y' + P(x)y = Q(x)[/tex] and there is a general way to solve such ODEs.
 
  • #3
cragar
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where Q(x)=0 and then use an integrating factor.
 
  • #4
elvishatcher
23
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Yep, seems like that oughta work
 
  • #5
cragar
2,552
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if you do that you get y=0.
 
  • #6
ehild
Homework Helper
15,543
1,915
It seems like you could solve this by first manipulating to get [tex]y' + \frac{Bx-C}{A}y = 0[/tex]

or [tex]y' = -\frac{Bx-C}{A}y [/tex]

which is separable. [tex]\frac{y'}{y} = -\frac{Bx-C}{A}[/tex].

ehild
 
  • #7
cragar
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wow can't believe I missed that , thanks for the help
ok so I would get
[itex] ln(y)= \frac{-1}{A}(\frac{Bx^2}{2}-Cx)+F [/itex]
F= integration constant
then I just raise each side to e and I will have y
 
  • #8
ehild
Homework Helper
15,543
1,915
Exactly. It will be a bit simpler if you eliminate the minus sign in front of the parentheses,

[tex]\ln(y)=\frac{1}{A}(Cx-B\frac{x^2}{2})+F[/tex]

ehild
 

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