Differential Equations Problem

Daweih
Messages
7
Reaction score
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?
 
Physics news on Phys.org
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.
 
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.
 
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:
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top