Separating Variables in Differential Equations: Solving dP/dt = P - P^2

In summary, the student is having trouble with integrating a function and realizes they made a mistake. They ask for help and were given a solution.
  • #1
bdh2991
103
0

Homework Statement


dP/dt=P-P^2


Homework Equations





The Attempt at a Solution



I know you can separate this and after i did that and did my partial fractions i got

t + C = ln(P) + ln(1-P) but i don't know what to do from here i figure you take the e of both sides at some point but i never ended up with the right andswer please help
 
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  • #2
bdh2991 said:

Homework Statement


dP/dt=P-P^2


Homework Equations





The Attempt at a Solution



I know you can separate this and after i did that and did my partial fractions i got

t + C = ln(P) + ln(1-P) but i don't know what to do from here i figure you take the e of both sides at some point but i never ended up with the right andswer please help

I think you have a sign error in your solution. You need to fix that and remember the formulas like ln(a)+ln(b) = ln(ab) and ln(a) - ln(b) = ln (a/b).
 
  • #3
i'm not seeing where the sign error is coming from...for my partial fractions i did

A/P + B/(1-P), then getting rid of the denominators i get A(1-P) + Bp = 1

solving for A and B i get A=1, B=1

so wouldn't it be ln(P) + ln(1-P)??

the only thing i could think is that it should be A(1-P) - BP

but i don't really understand how it could come out as a subtraction
 
  • #4
bdh2991 said:
i'm not seeing where the sign error is coming from...for my partial fractions i did

A/P + B/(1-P), then getting rid of the denominators i get A(1-P) + Bp = 1

solving for A and B i get A=1, B=1

so wouldn't it be ln(P) + ln(1-P)??

the only thing i could think is that it should be A(1-P) - BP

but i don't really understand how it could come out as a subtraction

Your partial fraction should be 1/P - 1/(P-1).
 
  • #5
bdh2991 said:
i'm not seeing where the sign error is coming from...for my partial fractions i did

A/P + B/(1-P), then getting rid of the denominators i get A(1-P) + Bp = 1

solving for A and B i get A=1, B=1

so wouldn't it be ln(P) + ln(1-P)??

Your sign is wrong on your integration of$$
\int\frac 1 {1-P}\, dP$$
 
  • #6
I see i was messing up my U-substitution, it's always the little things lol thanks for the help
 

Related to Separating Variables in Differential Equations: Solving dP/dt = P - P^2

1. What is separation of variables in differential equations?

Separation of variables is a method used to solve certain types of differential equations, where the equation can be separated into two or more simpler equations by writing the variables as separate functions and then integrating each side separately.

2. When is separation of variables used in solving differential equations?

Separation of variables is typically used when the differential equation is of the form y' = f(x)g(y), where the variables x and y can be separated into two different functions.

3. What is the general process for solving a differential equation using separation of variables?

The general process for solving a differential equation using separation of variables is to first write the equation in the form y' = f(x)g(y). Then, separate the variables by dividing both sides by g(y) and multiplying both sides by dx. Next, integrate each side separately, with respect to x and y. Finally, solve for y to obtain the solution to the differential equation.

4. Are there any limitations to using separation of variables in solving differential equations?

Yes, there are certain types of differential equations that cannot be solved using separation of variables, such as non-linear equations or equations with non-constant coefficients. It is important to check if the equation is suitable for separation of variables before using this method.

5. Can separation of variables be used for higher order differential equations?

Yes, separation of variables can also be used to solve higher order differential equations, where there are multiple variables involved. However, the process becomes more complex and may require additional techniques such as substitution or integration by parts.

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