1st order diff equ vs integral tables

  • Thread starter rppearso
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In summary: The residue theorem is an important tool there. Some integrals can be solved directly by differentiation under the integral sign.In summary, the conversation discusses the solution to two different types of equations: a first order differential equation and an integral. The differential equation solution has an extra constant (C1) that is not present in the integral solution. However, when taking into account constants of integration, the two equations are equivalent. The conversation also explores the use of integration constants and their role in solving complex integrals. The integral tables are created using a variety of methods, such as substitutions, integration by parts, and integration over contours in the complex plane.
  • #1
rppearso
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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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  • #2
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.
 
  • #3
smallphi said:
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 can't 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?
 
  • #4
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.
 
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  • #5
smallphi said:
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 can't 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.
 
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  • #6
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.
 

1. What is the difference between 1st order differential equations and integral tables?

1st order differential equations are mathematical equations that involve derivatives, while integral tables are collections of integrals that can be used to solve integration problems. In other words, differential equations involve finding the rate of change of a function, while integrals involve finding the original function given its rate of change.

2. When should I use a 1st order differential equation versus an integral table?

1st order differential equations are useful for solving problems involving rates of change, such as growth or decay problems. Integral tables are useful for finding the original function given its rate of change, and can be used to solve more complex integration problems.

3. Are there any limitations to using 1st order differential equations?

While 1st order differential equations can be used to solve many problems, they do have limitations. They are typically only useful for solving linear equations, and may not be able to accurately model certain non-linear systems.

4. How do I find the solutions to a 1st order differential equation?

To find solutions to a 1st order differential equation, you must first separate the variables and then use integration techniques to solve for the unknown function. The specific method used will depend on the type of equation and its initial conditions.

5. Can I use both 1st order differential equations and integral tables in the same problem?

Yes, it is possible to use both 1st order differential equations and integral tables in the same problem. For example, if you have a differential equation that involves a rate of change and need to find the original function, you can use an integral table to solve for the original function.

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