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avatar+1313 

dP/dt = 0.02P, P(0)=20. 

Please show workings with all steps. 

I have 0.02t^1.02/1.02=ln|P+C|, and the final solution is P=20e^0.02t. I cannot get from my solution to the final solution right?

 Aug 16, 2014

Best Answer 

 #4
avatar+128631 
+13

Because the initial condition gives P as some function of t. and we know that when t = 0, P = 20.

Think of P as being P(t)........kind of like  "f(x)" that you were used to dealing with in Algebra.

 

 Aug 16, 2014
 #1
avatar+4473 
+13

AzizHusain Aug 16, 2014
 #2
avatar+128631 
+13

Separating the variables and integrating, we have

∫1/P  dp  =  ∫.02 dt

ln P = .02t + C

This says that

e^(.02t + C) = P    .....or.....

(e^C)* e^(.02t) = P

And e^C is just a constant....we can just call it "C"

And we have P(0) = 20   .... so...

Ce^(.02(0)) = 20

 Therefore, C = 20

And our answer is

P = 20e^(.02t)

I hope this helps.....

 Aug 16, 2014
 #3
avatar+1313 
+8

What makes you solve it as P= and move everything else?

 Aug 16, 2014
 #4
avatar+128631 
+13
Best Answer

Because the initial condition gives P as some function of t. and we know that when t = 0, P = 20.

Think of P as being P(t)........kind of like  "f(x)" that you were used to dealing with in Algebra.

 

CPhill Aug 16, 2014
 #5
avatar+1313 
+8

It is easier also.

 Aug 16, 2014

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