Differential Equations Problem

In summary, the student had a question about a differential equation involving exponential growth and decay. They initially solved the problem correctly but did not include a negative sign in their final answer. After comparing their solution to the answer key, they realized their mistake and asked for clarification. Another student pointed out that they had not used the correct method of integration, which explained the discrepancy in the final answer.
  • #1
Daweih
7
0

Homework Statement


This is actually a question we went over in class, but I kind of spaced out when my teacher was explaining it. I have since solved it myself while I was reviewing for my calculus test, but when I compared my answer with the answer key my teacher provided to our class, I found out my answer was wrong. Here's the question.

"Write and solve the differential equation that models the statement in the following problem.

The rate of change of P with respect to t is proportional to 10-t."

Homework Equations


I should add that this is part of our chapter involving exponential growth and decay.


The Attempt at a Solution


Now, I understand how to do this particular problem. You essentially start off with the following differential equation:

dP/dt = K/(10-t)

You then multiply the dt over to isolate all the t variable expressions on the right side.

dP = [K/(10-t)]dt

Then you would take an indefinite integral of both sides respectively. The problem is, my teacher says that the answer should be:

P = -K*ln|10-t| + C

When I solved this on my own, I produced the same result essentially, just without a negative sign in front of the K constant. Am I missing something here? Could someone please explain to me why there should be a negative sign or did my teacher make a mistake?
 
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  • #2
Daweih said:

The Attempt at a Solution


Now, I understand how to do this particular problem. You essentially start off with the following differential equation:

dP/dt = K/(10-t)

You then multiply the dt over to isolate all the t variable expressions on the right side.

dP = [K/(10-t)]dt

Then you would take an indefinite integral of both sides respectively. The problem is, my teacher says that the answer should be:

P = -K*ln|10-t| + C

When I solved this on my own, I produced the same result essentially, just without a negative sign in front of the K constant. Am I missing something here? Could someone please explain to me why there should be a negative sign or did my teacher make a mistake?
Did you integrate the right side using u-substitution? After you use that method, the negative will appear in the answer.
 
  • #3
Ah, no. I didn't. I tried integrating using substitution just now, and it worked. >.<
Thank you for pointing that out to me. It makes sense now.
 
  • #4
Daweih said:

Homework Statement


This is actually a question we went over in class, but I kind of spaced out when my teacher was explaining it. I have since solved it myself while I was reviewing for my calculus test, but when I compared my answer with the answer key my teacher provided to our class, I found out my answer was wrong. Here's the question.

"Write and solve the differential equation that models the statement in the following problem.

The rate of change of P with respect to t is proportional to 10-t."

Homework Equations


I should add that this is part of our chapter involving exponential growth and decay.

The Attempt at a Solution


Now, I understand how to do this particular problem. You essentially start off with the following differential equation:

dP/dt = K/(10-t)

When you say ##y## is proportional to ##x## that means there is a constant ##k## such that ##y=kx##. So if the rate of change of ##P## is proportional to ##10-t##, then ##\frac{dP}{dt}= k(10-t)##. So you either are solving the problem wrong or stated the problem you are solving wrong.
 
Last edited:
  • #5
That's my mistake. I stated the problem incorrectly. It should say "inversely proportional". But regardless, eumyang pointed out what I was missing. I didn't use the substitution method to integrate the differential equation, so I never produced that negative sign.
 

FAQ: Differential Equations Problem

What are differential equations and why are they important in science?

Differential equations are mathematical equations that describe how a quantity changes over time. They are important in science because they allow us to model and predict the behavior of dynamic systems, such as population growth, chemical reactions, and electrical circuits.

How are differential equations solved?

Differential equations can be solved analytically using various methods, such as separation of variables, substitution, and integrating factors. They can also be solved numerically using computer algorithms, such as Euler's method or Runge-Kutta methods.

Are there any real-world applications of differential equations?

Yes, there are countless real-world applications of differential equations. Some examples include predicting the spread of diseases, designing control systems for airplanes and spacecraft, and modeling climate change.

What is the difference between ordinary and partial differential equations?

Ordinary differential equations involve one independent variable, such as time, and one or more dependent variables. Partial differential equations involve multiple independent variables, such as position and time, and multiple dependent variables.

Are there any limitations to using differential equations in science?

Yes, there are some limitations to using differential equations in science. They may not always accurately represent complex systems, and the solutions may not always be exact due to simplifications and assumptions made in the modeling process.

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