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

  • Thread starter Thread starter badtwistoffate
  • Start date Start date
Click For Summary
The discussion focuses on solving the ordinary differential equation y' = y(3 - y) using the method of separation of variables. The initial approach involves rewriting the equation and attempting to separate variables, but the user encounters confusion with the algebraic manipulation. After some back-and-forth, the correct method involves using partial fractions to integrate both sides, leading to the expression ln(y/(y-3)) = 3x + C. The conversation highlights the importance of understanding logarithmic identities and basic algebra in solving differential equations. Overall, the thread emphasizes the step-by-step process of finding the general solution while addressing common misunderstandings.
badtwistoffate
Messages
81
Reaction score
0
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.
 
Last edited:
Physics news on Phys.org
\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
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
49
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
Replies
3
Views
2K
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K