Question

general solution of (dP/dr)= P^2

Answer 1

use power rule
I am thinking that is perimeter and radius?

Answer 2

I think the answer is P=(Ce^r)^(1/2)
but I just need to be sure

Answer 3

You can check your answer by taking the derivative and seeing if your equation substituting your solution gets you an undeniably true equation (like 0 = 0 is always true).

Answer 4

separate the variables and integrate
\[\int\limits \frac{ dP }{ P^2 }=\int\limits dr\]
\[\int\limits P ^{-2}dP=r+c,\frac{ -1 }{P }=r+c\]

Answer 5

okay, this is interest rate

Answer 6

right I did that and I got \[\ln (P^2)=r\] then I have to solve for P so for that I got \[P^2= e^{r+c}\]

Answer 7

which is also \[P^2= (e ^{r})(e^c)\]

Answer 8

then (because of my teacher) I have to substitute C for e^c that makes the whole thing

if
\[ C=e^c \]
then
\[P^2=Ce^r\]

Answer 9

Can we back up? (I can see where you are going to get to the solution you found).
Where did you get: \( \ln \left( P^2 \right) = r \) from?
From separating the variables and integrating?

Answer 10

yes

Answer 11

actually ln(P^2)=r+c

Answer 12

my bad. I always forget to add in the c

Answer 13

Ohh, I can see what you did.
Careful, this is not always true:
\( \displaystyle \int \frac{1}{x^n} \ dx = \ln x^n + c \)

In fact, it only works for n=1.
\( \displaystyle \int \frac{1}{x} \ dx = \ln x^1 + c \)
In any other case, we use power rule on a negative power:
\( \displaystyle \int \frac{1}{x^n} \ dx = \int x^{-n} \ dx = \frac{x^{-n+1}}{-n+1} + c \)

Notice that this latter integral /only/ fails on n=1, where the denominator is 0.

Answer 14

Errrr... so its \[(P^-2)/2= e ^{r+c}\]

Answer 15

no its not even that its ((P^-2)/2)=r+c

Answer 16

Allmosttt, The power was originally -2. We have to add 1 to the power and divide that off.

Answer 17

Sorry please be patient with me. So it's (P^-1/-1)

Answer 18

It's fine. :)
and yes, there you go!
P^(-1) / -1 = r + c is good now.

Answer 19

aww I was so excited I thought I had the right answer. Now I have to rework the whole thing. Hang on.

Answer 20

So the general solution of (dP/dr)=P^2
is
P= 1/(r+c)

Answer 21

you missed a very small detail from P^(-1) / -1 = r + c

Answer 22

the negative

Answer 23

damn.

Answer 24

ok its -1/(r+c)

Answer 25

yes. easy to miss but important. Don't feel bad though, we always make mistakes, and it is good to learn from those happy little accidents. :p

Answer 26

Okay thanks! You have been so helpful otherwise I would have missed 20 points on a take home math test!

Answer 27

if you are ever uncertain, too, we always have the ability to check just by substituting back into the original equation.
finding P^2 and dP/dr and substituting back into the original equation will always create a true equation in the end if P is correct.

Answer 28

Okay thanks again!

Answer 29

glad to help! :)