How Can I Solve This Challenging Bernoulli Equation?

  • Thread starter cragar
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In summary, the equation given can be solved using undetermined coefficients after manipulating it to the form of a separable ODE. The solution involves taking the natural log of y and setting it equal to a constant plus a function of x. This can then be solved for y by raising both sides to the power of e.
  • #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.
 
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  • #2
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
where Q(x)=0 and then use an integrating factor.
 
  • #4
Yep, seems like that oughta work
 
  • #5
if you do that you get y=0.
 
  • #6
elvishatcher said:
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
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
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
 

FAQ: How Can I Solve This Challenging Bernoulli Equation?

What is the purpose of solving Ay'+Bxy=Cy: f(x)?

The purpose of solving Ay'+Bxy=Cy: f(x) is to find a solution for the differential equation, which represents the relationship between a function and its derivatives. This can be used to model real-world phenomena, such as population growth or the motion of objects.

How do you solve Ay'+Bxy=Cy: f(x)?

To solve Ay'+Bxy=Cy: f(x), you first need to separate variables and integrate both sides of the equation. Then, you can use algebraic manipulation and the initial conditions to find the particular solution. If there is a constant term in the equation, you may also need to use the method of undetermined coefficients.

What are the initial conditions in solving Ay'+Bxy=Cy: f(x)?

The initial conditions are the values of the function and its derivatives at a specific point. They are necessary to find the particular solution of the differential equation and can be provided in the form of given values or a graph.

Can Ay'+Bxy=Cy: f(x) have multiple solutions?

Yes, Ay'+Bxy=Cy: f(x) can have multiple solutions. This is because a differential equation represents a family of functions, and the initial conditions are used to find a specific member of that family. Therefore, different initial conditions can lead to different solutions.

What are some applications of solving Ay'+Bxy=Cy: f(x)?

Solving Ay'+Bxy=Cy: f(x) has many applications in various fields such as physics, engineering, biology, and economics. It can be used to model exponential growth and decay, motion under the influence of forces, and other dynamic processes. It is also essential in solving problems involving rates of change and optimization.

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