1st order diff equ vs integral tables


by rppearso
Tags: diff, integral, order, tables
rppearso
rppearso is offline
#1
Sep22-08, 08:55 PM
P: 95
I am at a dilemma trying to solve a simple first order differential equation of the form;

dT/dt + C1*T = C2 the solution to this differential equation is T = C2/C1 + exp(-C1t)

Which is equivalent to t= 1/-C1*ln(T+C2/C1)

The integral table states that the solution to an integral of the form:

dT/(aT+b) = 1/a*ln(a*T+b)

Both of these equations assume no constants of integration,

There is an extra C1 in the differential equation solution that is not in the integral.
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smallphi
smallphi is offline
#2
Sep22-08, 10:11 PM
P: 443
When you take into account the constants of integration, the two formulas would be the same because constants of integration can absorb the differences.

For example the two expression below are exactly the same:

k1*exp(x)
exp(x+k2)

where k1, k2 are integration constants. The connection is k1=exp(k2). Any function of an integration constant is also an integration constant. You can reformulate integration constants that way to check if two seemingly different expressions are the same.
rppearso
rppearso is offline
#3
Sep23-08, 12:48 AM
P: 95
Quote Quote by smallphi View Post
When you take into account the constants of integration, the two formulas would be the same because constants of integration can absorb the differences.

For example the two expression below are exactly the same:

k1*exp(x)
exp(x+k2)

where k1, k2 are integration constants. The connection is k1=exp(k2). Any function of an integration constant is also an integration constant. You can reformulate integration constants that way to check if two seemingly different expressions are the same.
The integration constant for the differential equation would be in front of the exponential and when you rearanged the equation you would end up with;

ln((-C1*T+C2)/C*C1) and with the integral table you end up with ln(-C1*T+C2) + C even with integration constants taken into account I cant seem to properly shake the C1 out of my equation to match the integral tables. The differential equation and integral table should match.

Another question, do they use differential equation methods to solve complex integrals in the integral tables like use of integrating factors?

smallphi
smallphi is offline
#4
Sep23-08, 09:24 AM
P: 443

1st order diff equ vs integral tables


ln((-C1*T+C2)/C*C1) = ln(-C1*T+C2) +ln(C1/C)

Now ln(C1/C) is a function of integration constant that doesn't contain the variable, T in this case, so it can be considered integration constant too. That is confirmed by the fact that the value of ln(C1/C) doesn't depend on T and is arbitrary due to the presence of C so it behaves exactly like integration constant. Denote it by ln(C1/C) -> C and you get the second expression.
rppearso
rppearso is offline
#5
Sep23-08, 11:35 AM
P: 95
Quote Quote by smallphi View Post
ln((-C1*T+C2)/C*C1) = ln(-C1*T+C2) +ln(C1/C)

Now ln(C1/C) is a function of integration constant that doesn't contain the variable, T in this case, so it can be considered integration constant too. That is confirmed by the fact that the value of ln(C1/C) doesn't depend on T and is arbitrary due to the presence of C so it behaves exactly like integration constant. Denote it by ln(C1/C) -> C and you get the second expression.
I cant believe I forgot my log rules, actually wouldent it be ln(-C1*T+C2) - ln(C1*C). Either way it answers my question.

Thank you,
Ron

BTW are the integral tables created through differential equation methods of solving. Some of thoes complex integrals are not straight forward to solve using integration rules.
smallphi
smallphi is offline
#6
Sep23-08, 12:50 PM
P: 443
Indefinite integrals are solved by substitutions, change of integration variables, integration by parts etc. the usual tricks in a calculus book. The substitutions/change of variables are not always obvious.

Some definite integrals on the real axis are solved by integration over contours in the complex plane.


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