How Can I Solve This Challenging Bernoulli Equation?

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Homework Help Overview

The discussion revolves around a Bernoulli equation of the form Ay' + Bxy = Cy, where A, B, and C are real constants. Participants explore methods to manipulate the equation and discuss potential approaches to finding a solution.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants consider using an integrating factor and manipulating the equation into a standard form for first-order ODEs. There are discussions about the implications of the right side being a function of x and the potential use of undetermined coefficients.

Discussion Status

Several participants have provided insights into manipulating the equation, suggesting it can be expressed in a separable form. There is acknowledgment of the validity of these approaches, but no consensus on a final method has been reached.

Contextual Notes

Participants are navigating the complexities of the equation's structure and the implications of the constants involved. There is an ongoing exploration of assumptions regarding the form of the solution and the integration process.

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.
 
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It seems like you could solve this by first manipulating to get y' + \frac{Bx-C}{A}y = 0 at which point you now have an ODE of the form y' + P(x)y = Q(x) and there is a general way to solve such ODEs.
 
where Q(x)=0 and then use an integrating factor.
 
Yep, seems like that oughta work
 
if you do that you get y=0.
 
elvishatcher said:
It seems like you could solve this by first manipulating to get y' + \frac{Bx-C}{A}y = 0

or y' = -\frac{Bx-C}{A}y

which is separable. \frac{y'}{y} = -\frac{Bx-C}{A}.

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

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

ehild
 

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