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Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 9, Problem 9.2.28a

28. Derivation of Equation (7) in Example 4


a. Show that the solution of the equation
di /dt + R/Li = V/L
is
i = V/R + Cexp(-(R/L)i) .

Verified step by step guidance
1
Identify the given differential equation: \(\frac{di}{dt} + \frac{R}{L} i = \frac{V}{L}\), which is a first-order linear ordinary differential equation in the variable \(i(t)\).
Rewrite the equation in the standard linear form: \(\frac{di}{dt} + P(t) i = Q(t)\), where \(P(t) = \frac{R}{L}\) and \(Q(t) = \frac{V}{L}\), both constants in this case.
Find the integrating factor \(\mu(t)\) using the formula \(\mu(t) = e^{\int P(t) dt} = e^{\int \frac{R}{L} dt} = e^{\frac{R}{L} t}\).
Multiply both sides of the differential equation by the integrating factor to get: \(e^{\frac{R}{L} t} \frac{di}{dt} + \frac{R}{L} e^{\frac{R}{L} t} i = \frac{V}{L} e^{\frac{R}{L} t}\), which simplifies to \(\frac{d}{dt} \left( e^{\frac{R}{L} t} i \right) = \frac{V}{L} e^{\frac{R}{L} t}\).
Integrate both sides with respect to \(t\): \(\int \frac{d}{dt} \left( e^{\frac{R}{L} t} i \right) dt = \int \frac{V}{L} e^{\frac{R}{L} t} dt\), then solve for \(i(t)\) by dividing by \(e^{\frac{R}{L} t}\) and include the constant of integration \(C\) to obtain the general solution.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

First-Order Linear Differential Equations

This type of differential equation has the form dy/dt + P(t)y = Q(t). Solving it involves finding an integrating factor to simplify the equation and integrate both sides. Recognizing the given equation as first-order linear is essential to apply the correct solution method.
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Integrating Factor Method

The integrating factor is a function, usually e^(∫P(t)dt), used to multiply the entire differential equation to make the left side an exact derivative. This technique transforms the equation into a form that can be integrated directly, enabling the solution for the unknown function.
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Solving for the Constant of Integration

After integrating, the solution includes an arbitrary constant C representing initial conditions. Understanding how to incorporate this constant and interpret its role in the general solution is crucial for expressing the complete family of solutions to the differential equation.
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