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In my textbook, the authors tried to solve the differential equation: $dy/dt+ay=g(t)$, where $a$ is a constant and $g(t)$ is a function. Why the authors of my textbook tend to leave the answer in terms of definite integral instead of indefinite integral if the indefinite integral cannot be evaluated in terms of elementary functions (the authors leave the answer as $y=e^{-at}\int_{t_0}^t e^{as} g(s) ds + ce^{-at}$ (by using s as a dummy variable of integration, c as an arbitrary constant, $t_0$ as a convenient lower limit of integration) instead of $y=e^{-at}\int e^{at} g(t) dt + ce^{-at}$ if $\int e^{at} g(t) dt$ cannot be evaluated in terms of elementary functions). Please refer to the text:

http://issuu.com/wiley_publishing/docs/boyce_elementary_differential_equations_10e_sample?e=1085234/2816160

On pages: 5-6 (34-35) (the numbers in the parentheses are the actual page numbers), the authors wrote "... For many simple functions g(t), we can evaluate the integral in Eq. (23) and express the solution y in terms of elementary functions, as in Example 2. However, for more complicated functions g(t), it is necessary to leave the solution in integral form. In this case $y=e^{-at}\int_{t_0}^t e^{as} g(s) ds +ce^{-at}$..."

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The answer in part is it makes it incorporating the initial condition easier. For instance, if $y_0 = y(t_0) = ce^{-at_0}$ then it's clean and easy to solve for $c$.

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  • $\begingroup$ The authors did not mentioned that the problem is an initial-value problem. No initial condition is given by the authors. $\endgroup$ – shuxue Dec 14 '14 at 17:08
  • $\begingroup$ None the less, it does make writing down the explicit solution to the IVP easy. Maybe others can propose other reasons. $\endgroup$ – Simon S Dec 14 '14 at 17:09
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This is an additional reason to the one in Simon's comment.

The solution is the sum of

  1. The general solution $c\,e^{-at}$
  2. A particular solution $e^{-at}\int_{t_0}^t e^{as} g(s)\,ds$

Writing it as a definite integral you just choose a particular solution.

The solution could also be written as $$ e^{-at}\int e^{as} g(s)\,ds, $$ since $$ \int e^{as} g(s)\,ds = \int_{t_0}^t e^{as} g(s)\,ds+c. $$

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  • $\begingroup$ Why the authors prefer to leave the answer as $y=e^{-at}\int_{t_0}^t e^{as} g(s) ds + ce^{-at}$ instead of $y=e^{-at}\int e^{at} g(t) dt + ce^{-at}$ ? $\endgroup$ – shuxue Dec 14 '14 at 19:21
  • $\begingroup$ You should ask them. But they would never use $y=e^{-at}\int_{t_0}^t e^{as} g(s) ds +ce^{-at}$, since the indefinite integral already has a constant of integration in it. $\endgroup$ – Julián Aguirre Dec 14 '14 at 19:55

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