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

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In summary, the student is having difficulty solving an ordinary differential equation. They start by finding the general solution by multiplying the y through so it becomes: y'=3y-y2 and do the separation of variables by dividing 3y-y2 to the other side so you get: dy/3y-y^2=0. They get stuck when they try to solve for y and are unable to do so without help from the instructor.
  • #1
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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  • #2
[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.
 
  • #3
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)
 
  • #4
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. :uhh:
 
  • #5
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?
 
  • #6
You have the same thing. Just change your - sign into an exponent and multiply both sides by 3.
 
  • #7
how do you change a minus sigh into a exponent, I am lost when you said that.
 
  • #8
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]
 
  • #9
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
 

1. What is the "Simple Method for Solving y'=y(3-y) ODE"?

The "Simple Method for Solving y'=y(3-y) ODE" is a method for finding the general solution of the first-order ordinary differential equation (ODE) with the form y' = y(3-y). This method involves separating the variables and integrating both sides of the equation.

2. How do you solve y'=y(3-y) ODE using the Simple Method?

To solve y'=y(3-y) ODE using the Simple Method, we first rewrite the equation as y' = 3y - y^2. Then, we separate the variables by dividing both sides by (3y - y^2). This gives us the equation (1/y - 1/(3-y))dy = dx. We can then integrate both sides to get the general solution.

3. What are the steps involved in the Simple Method for Solving y'=y(3-y) ODE?

The steps involved in the Simple Method for Solving y'=y(3-y) ODE are: 1) Rewrite the equation in the form y' = 3y - y^2; 2) Separate the variables by dividing both sides by (3y - y^2); 3) Integrate both sides to get the general solution; 4) Solve for any constants using initial conditions, if given.

4. What are the advantages of using the Simple Method for Solving y'=y(3-y) ODE?

One advantage of using the Simple Method for Solving y'=y(3-y) ODE is that it is a straightforward and systematic approach that can be easily applied to other first-order ODEs with similar forms. Additionally, it does not require any special techniques or knowledge of advanced mathematics.

5. Are there any limitations to the Simple Method for Solving y'=y(3-y) ODE?

The Simple Method for Solving y'=y(3-y) ODE may not work for more complex first-order ODEs that do not have the form y' = y(3-y). In these cases, other methods, such as separation of variables or integrating factors, may be more appropriate. Additionally, the Simple Method may not yield a solution for certain initial conditions, in which case a more specialized approach may be needed.

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