Solving an Initial Value Problem for x(t=ln2): Step-by-Step Solution

In summary, an Initial Value Problem (IVP) is a type of differential equation that involves finding the solution to an equation given a specific initial value. The main difference between an IVP and a Boundary Value Problem (BVP) is that an IVP requires a specific initial value, while a BVP requires conditions to be met at both the beginning and end of the interval in which the solution is sought. The steps to solving an IVP typically involve transforming the equation into a simpler form, finding the general solution, and then using the initial value to determine the specific solution. IVPs are important in science because they are used to model real-world phenomena and make predictions about their behavior. Some common methods for solving IVPs include
  • #1
Drakkith
Mentor
23,093
7,502

Homework Statement


Solve the initial value problem:
##\frac{dx}{dt} = x(2-x)##, ##x(0) = 1##
for ##x(t=ln2)##.

Homework Equations

The Attempt at a Solution



I moved the right side to the left and multiplied both sides by dt to get:
##\frac{dx}{x(2-x)} = dt##

Integrating gave me:
##\frac{ln|x|}{2} - \frac{ln|x-2|}{2} = t+C##

Then:
##ln|x| - ln|x-2| = 2t + 2C##
##ln|\frac{x}{x-2}| = 2t + 2C##
##\frac{x}{x-2} = e^{2t}e^{2c}##
##\frac{1}{1-\frac{2}{x}} = Ke^{2t}##

Manipulating for a while, I end up with:
##x=\frac{2}{Ke^{2t}}##

Since ##x(0) = 1##, I set the left side to 1 and solve for K, winding up with ##k=3##.
However, when trying to solve for ##x(ln2)## I end up with ##\frac{2}{11}##, which isn't one of my possible answers.

Does my process look even remotely correct?
 
Physics news on Phys.org
  • #2
Drakkith said:

Homework Statement


Solve the initial value problem:
##\frac{dx}{dt} = x(2-x)##, ##x(0) = 1##
for ##x(t=ln2)##.

Homework Equations

The Attempt at a Solution



I moved the right side to the left and multiplied both sides by dt to get:
##\frac{dx}{x(2-x)} = dt##

Integrating gave me:
##\frac{ln|x|}{2} - \frac{ln|x-2|}{2} = t+C##

Then:
##ln|x| - ln|x-2| = 2t + 2C##
##ln|\frac{x}{x-2}| = 2t + 2C##
##\frac{x}{x-2} = e^{2t}e^{2c}##
##\frac{1}{1-\frac{2}{x}} = Ke^{2t}##
So you could say ##\frac x {x-2} = Ke^{2t}## Put ##t=0,~x=1## there to get ##\frac x {x-2} = -e^{2t}##. Now put ##t=\ln 2## in that, then solve for ##x##. See if that fixes it.
 
  • Like
Likes Drakkith
  • #3
Drakkith said:

Homework Statement



****************.

##ln|x| - ln|x-2| = 2t + 2C##
##ln|\frac{x}{x-2}| = 2t + 2C##
##\frac{x}{x-2} = e^{2t}e^{2c}##
##\frac{1}{1-\frac{2}{x}} = Ke^{2t}##

For ##t## near 0 you have ##x## near 1, and so ##x/(x-2) = e^{2C} e^{2t}## is impossible unless you let ##C## be a complex number. However, ##\ln(|x|/|x-2|) = \ln(x/(2-x))##, and so having ##x/(2-x) = e^{2C} e^{2t}## is OK. Of course, when you re-wrote ##e^{2C}## as ##K## and then forgot the origin of ##K##, you were then able to have a valid formula!
 
  • Like
Likes Drakkith
  • #4
Drakkith said:
##ln|\frac{x}{x-2}| = 2t + 2C##
##\frac{x}{x-2} = e^{2t}e^{2c}##
##\frac{1}{1-\frac{2}{x}} = Ke^{2t}##

Ray Vickson said:
For ##t## near 0 you have ##x## near 1, and so ##x/(x-2) = e^{2C} e^{2t}## is impossible unless you let ##C## be a complex number. However, ##\ln(|x|/|x-2|) = \ln(x/(2-x))##, and so having ##x/(2-x) = e^{2C} e^{2t}## is OK. Of course, when you re-wrote ##e^{2C}## as ##K## and then forgot the origin of ##K##, you were then able to have a valid formula!

@Drakkith: Ray is quite correct and I didn't notice you had been sloppy with the absolute value signs. A better way for you to have written it would be like this:$$
ln\left | \frac{x}{x-2}\right | = 2t + 2C$$ $$
\left|\frac{x}{x-2}\right | =e^{2t + 2C}=e^{2C}e^{2t}$$ $$
\frac{x}{x-2} =\pm e^{2C}e^{2t} = Ke^{2t}$$
 
  • Like
Likes Drakkith
  • #5
LCKurtz said:
So you could say ##\frac x {x-2} = Ke^{2t}## Put ##t=0,~x=1## there to get ##\frac x {x-2} = -e^{2t}##. Now put ##t=\ln 2## in that, then solve for ##x##. See if that fixes it.

Yes, that seemed to work. I got x=8/5, which is one of the possible answers. Thanks guys!
 

FAQ: Solving an Initial Value Problem for x(t=ln2): Step-by-Step Solution

What is an Initial Value Problem?

An Initial Value Problem (IVP) is a type of differential equation that involves finding the solution to an equation given a specific initial value. The initial value is typically given as a point on the graph of the solution.

What is the difference between an IVP and a Boundary Value Problem?

The main difference between an IVP and a Boundary Value Problem (BVP) is that an IVP requires a specific initial value, while a BVP requires conditions to be met at both the beginning and end of the interval in which the solution is sought.

What are the steps to solving an IVP?

The steps to solving an IVP typically involve transforming the equation into a simpler form, finding the general solution, and then using the initial value to determine the specific solution. These steps may vary depending on the type of differential equation and the methods used for solving it.

Why are IVPs important in science?

IVPs are important in science because they are used to model real-world phenomena and make predictions about their behavior. Many natural processes and systems can be described using differential equations, and solving IVPs allows scientists to understand and predict their behavior.

What are some common methods for solving IVPs?

Some common methods for solving IVPs include separation of variables, substitution, and use of integrating factors. Other techniques such as Laplace transforms and numerical methods may also be used depending on the complexity of the equation and the desired level of accuracy.

Back
Top