# Differential Equation Initial Value Problem

Staff Emeritus

## Homework Statement

A Solve the following initial value problem:

##\frac{dx}{dt}=-x(1-x)##
##x(0)=\frac{3}{2}##

B. At what finite time does ##x→∞##

## The Attempt at a Solution

##\frac{dx}{dt}=x(x-1)##
##\frac{dx}{x(x-1)}=dt##
Partial fractions: ##dx(\frac{-1}{x}-\frac{1}{x-1})=dt##
Integrating both sides: ##ln|\frac{1}{x}|-ln|x-1|=t+c##
##ln|\frac{1}{x(x-1)}|=t+c##
e to the power of both sides and taking the constant ##e^c## as A: ##\frac{1}{x(x-1)}=Ae^t##
Plugging in the initial value gives me ##A=\frac{4}{3}##

My final equation is: ##\frac{1}{x(x-1)} = \frac{4}{3}e^t##

What I don't understand is how ##x## is related to ##t## and how to figure out at what time x goes to infinity. Offhand I don't see any way for X to increase to infinity since t is an exponent of e, unless t goes to negative infinity.

LCKurtz
Homework Helper
Gold Member

## Homework Statement

A Solve the following initial value problem:

##\frac{dx}{dt}=-x(1-x)##
##x(0)=\frac{3}{2}##

B. At what finite time does ##x→∞##

## The Attempt at a Solution

##\frac{dx}{dt}=x(x-1)##
##\frac{dx}{x(x-1)}=dt##
Partial fractions: ##dx(\frac{-1}{x}-\frac{1}{x-1})=dt##

Those fractions don't add up to what you started with.

Integrating both sides: ##ln|\frac{1}{x}|-ln|x-1|=t+c##
##ln|\frac{1}{x(x-1)}|=t+c##
e to the power of both sides and taking the constant ##e^c## as A: ##\frac{1}{x(x-1)}=Ae^t##
Plugging in the initial value gives me ##A=\frac{4}{3}##

My final equation is: ##\frac{1}{x(x-1)} = \frac{4}{3}e^t##

What I don't understand is how ##x## is related to ##t## and how to figure out at what time x goes to infinity. Offhand I don't see any way for X to increase to infinity since t is an exponent of e, unless t goes to negative infinity.

You can always solve for ##x## in terms of ##t##.

cnh1995
Homework Helper
Gold Member
What I don't understand is how xxx is related to ttt and how to figure out at what time x goes to infinity. Offhand I don't see any way for X to increase to infinity since t is an exponent of e, unless t goes to negative infinity.
Fix the partial fractions. A sign error can turn division into multiplication when you take logarithm.

Staff Emeritus
Those fractions don't add up to what you started with.

Gah! That's no good! Looks like I dropped a negative sign.

New fractions: ##\frac{1}{x-1}-\frac{1}{x}##
That means the problem becomes: ##\frac{x-1}{x}=Ae^t##
##1-\frac{1}{x}=Ae^t##
##1-\frac{1}{\frac{3}{2}}=A##
##A=\frac{1}{3}##

That puts the final equation as: ##x=\frac{3}{3-e^t}##
That means that x goes to infinity at t=3.

You can always solve for ##x## in terms of ##t##.

Indeed. It's just more difficult when your equation isn't correct in the first place.

LCKurtz
Homework Helper
Gold Member
Gah! That's no good! Looks like I dropped a negative sign.

New fractions: ##\frac{1}{x-1}-\frac{1}{x}##
That means the problem becomes: ##\frac{x-1}{x}=Ae^t##
##1-\frac{1}{x}=Ae^t##
##1-\frac{1}{\frac{3}{2}}=A##
##A=\frac{1}{3}##

That puts the final equation as: ##x=\frac{3}{3-e^t}##
That means that x goes to infinity at t=3.
Hopefully, that ##t=3## is just a typo.

Staff Emeritus
Hopefully, that ##t=3## is just a typo.

I take it you mean that it should be ##t→3##?

LCKurtz
Homework Helper
Gold Member
No. ##t=3## is not the correct answer.

cnh1995
Homework Helper
Gold Member
I take it you mean that it should be ##t→3##?
The entire denominator should tend to zero.

Staff Emeritus
Christ, please tell me it's when ##e^t→3##...

Staff Emeritus
The entire denominator should tend to zero.

Indeed. I promise I'm not as stupid as I appear... maybe.

cnh1995
Homework Helper
Gold Member
Christ, please tell me it's when ##e^t→3##...
Right.

LCKurtz
Homework Helper
Gold Member
What value of ##t## gives ##e^t = 3##?

Staff Emeritus
What value of ##t## gives ##e^t = 3##?

I believe it's ##ln(3)##. But my basic math has been so bad for this question that I don't know if I'd double down on that bet.

LCKurtz
Homework Helper
Gold Member
##\ln 3## is correct.

cnh1995
Homework Helper
Gold Member
I promise I'm not as stupid as I appear..
Relax! That's just a silly mistake..No big deal!
I believe it's ln(3)ln(3)ln(3).
Yes.

Staff Emeritus