Solving differential equations with definite integrals?

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SUMMARY

This discussion focuses on solving differential equations using definite integrals, specifically in the context of a calculus-based physics course. The equation m dv/dt = -A v² is analyzed, demonstrating the integration of both sides using definite integrals, leading to the result (m/A)(1/v_f - 1/v_i) = t_f. The participant questions the merits of using definite integrals versus indefinite integrals, noting that both methods yield equivalent results but may differ in clarity and complexity for initial-value problems. The conclusion emphasizes that while both approaches are valid, the choice may depend on the complexity of the problem and the clarity of understanding required.

PREREQUISITES
  • Understanding of differential equations, specifically first-order equations
  • Familiarity with definite and indefinite integrals
  • Knowledge of the Fundamental Theorem of Calculus (FTC)
  • Basic concepts of initial-value problems in calculus
NEXT STEPS
  • Explore the application of the Fundamental Theorem of Calculus in solving differential equations
  • Study the differences between definite and indefinite integrals in various contexts
  • Learn about initial-value problems and their significance in differential equations
  • Investigate higher-order differential equations and their solutions
USEFUL FOR

Students in calculus-based physics courses, educators teaching differential equations, and anyone interested in the practical applications of integration techniques in solving initial-value problems.

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I'm taking a calculus-based physics course, and we were solving a simple differential equation for a model of drag by separating variables: (where A is some arbitrary constant)

m \frac {dv} {dt} = -A v^2
- \frac {m} {A} \frac {dv} {v^2} = dt

My teacher then integrates both sides, but unlike in my calculus class, he uses definite integrals:

- \frac {m} {A} \int_{v_i}^{v_f} v^{-2} dv = \int_{t_i}^{t_f} dt

Initial time will be zero for simplicity, so using the FTC:

\frac {m} {A} (\frac {1} {v_f} - \frac {1} {v_i}) = t_f

I understand how to solve it as is done from my calculus class, using indefinite integrals and solving for the constant of integration C = \frac {m} {A v_i}, which gives an equivalent result.

So what's the merit of using one method as opposed to another? It seems to me like using definite integrals is quicker.

So if I do physics problems this way, why should/shouldn't I do the initial-value problems I get in calculus using definite integrals like this? Like when it asks to solve dy/dx = 3y, y(2) = 5 or such, what's wrong with doing \frac {1} {3} \int_{5}^{y_f} y^{-1} dy = \int_{2}^{x_f} dx? Both methods give the same result, and again, it seems quicker to do this the definite integral way rather than solving for C.

I'm guessing that it might be harder to understand what's going on as things get more complex, or something? Or I guess using subscripts on variables to really indicate evaluating the function for the independent variable is problematic?

(Whew, sorry for the length, but this has been bugging me for a while.)
 
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It really doesn't matter. What matters is that you are given a differential equation with an initial condition. From the differential equation you can find the general solution, and then somehow you have to impose the initial condition. You can do this after writing down the general solution (by first doing an indefinite integral and then solving for the unknown integration constant) or you solve the equation with the correct starting conditions right away. Of course they (should) give the same results in the end. It works the same way for higher order equations, e.g.
\frac{d^2x}{dt^2} = a
(Which I won't work out because I'm not really comfortable working with the infinitesimals as you did above.)
 

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