Simple Method for Solving y'=y(3-y) ODE

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Homework Help Overview

The discussion revolves around solving the ordinary differential equation y' = y(3 - y). Participants are exploring methods to find the general solution, primarily focusing on separation of variables and partial fractions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to separate variables but expresses confusion about the steps. Other participants suggest using partial fractions and integrating both sides, while some question the algebraic manipulations involved.

Discussion Status

Participants are actively engaging with the problem, offering various methods and questioning each other's approaches. There is a mix of guidance and critique, with some participants providing insights into logarithmic identities and algebraic properties.

Contextual Notes

Some participants express concern over the algebraic skills necessary for solving the differential equation, indicating a potential gap in foundational knowledge. The discussion reflects a range of interpretations and methods without reaching a consensus on the best approach.

badtwistoffate
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just started ordinary differential equation class, and i have to find the general soln of the eqn: y'=y(3-y)... just started so i only know of the method of separation of varibles.

Thought of multiplying the y through so it becomes:
y'=3y-y^2 then do..

dy/dx=3y-y^2 then... I am stuck... I tried doing separation of varibles by

Dividing 3y-y^2 to the other side so you get:

dy/3y-y^2=0? But that doesn't seem right.
 
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[tex]\frac{dy}{dx} = y(3-y)[/tex]

[tex]\frac{dy}{y(3-y)} = dx[/tex]

[tex]\int \frac{dy}{y(3-y)} = \int dx[/tex]

Break up the left-hand side by the method of partial fractions.
 
after partial fractions and integrating both sides:

ln(y) - ln(y-3) = 3 x + Cby logarithmic identities:

ln(y/(y-3)) = 3 x + CTaking the exponential of both sides:

y/(y-3) = C1 e^(3 x)

Where C1 = e^CSolving for y:

y = 3 C1 e^(3 x)/(C1 e^(3 x)-1)
 
Well, geez... why don't you just do the whole thing for him? Oh wait, you just did.

Way to encourage the joy of discovery. :rolleyes:
 
wurth_skidder_23 said:
after partial fractions and integrating both sides:

ln(y) - ln(y-3) = 3 x + C


by logarithmic identities:

ln(y/(y-3)) = 3 x + C


Taking the exponential of both sides:

y/(y-3) = C1 e^(3 x)

Where C1 = e^C


Solving for y:

y = 3 C1 e^(3 x)/(C1 e^(3 x)-1)


After my partial fraction decomposition I got:

-ln((y-3)/y)/3=x+c?
 
You have the same thing. Just change your - sign into an exponent and multiply both sides by 3.
 
how do you change a minus sigh into a exponent, I am lost when you said that.
 
Use the property of logarithms: [tex]r log b = log b^{r}[/tex]

where [tex]r = -1[/tex]

and [tex]b = \frac{y-3}{y}[/tex]
 
I'm afraid you are going to find differential equations extremely difficult if you cannot do basic algebra!
Starting from dy/dx=3y-y2 and dividing both sides by 3y- y2, you do NOT get dy/(3y- y2)= 0 any more than dividing both sides of xy= 3 by 3 would give xy/3= 0.
 
  • #10
It was a brain fart ease up on me
 

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