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

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SUMMARY

The discussion focuses on solving the ordinary differential equation (ODE) y' = y(3 - y) using the method of separation of variables. The correct approach involves rewriting the equation as dy/(y(3 - y)) = dx and applying partial fraction decomposition. After integrating both sides, the solution is derived as y = 3C1e^(3x)/(C1e^(3x) - 1), where C1 = e^C. Participants emphasize the importance of algebraic manipulation and logarithmic identities in solving ODEs.

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  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with separation of variables method
  • Knowledge of partial fraction decomposition
  • Basic logarithmic identities and properties
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  • Study the method of separation of variables in depth
  • Learn about partial fraction decomposition techniques
  • Explore logarithmic identities and their applications in calculus
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Students in introductory differential equations courses, educators teaching ODEs, and anyone seeking to strengthen their algebraic manipulation skills in calculus.

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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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\frac{dy}{dx} = y(3-y)

\frac{dy}{y(3-y)} = dx

\int \frac{dy}{y(3-y)} = \int dx

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: r log b = log b^{r}

where r = -1

and b = \frac{y-3}{y}
 
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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