Question

Use an integrating factor to obtain the general solution of \(i R+L \frac{\mathrm{d} i}{\mathrm{~d} t}=\sin \omega t\) where \(R, L\) and \(\omega\) are constants.

Short answer

#Short Answer: The general solution of the given ODE when multiplied with the integrating factor can be obtained by following these steps: 1. Rewrite the ODE in standard form: \(\frac{di}{dt} + \frac{R}{L}i = \frac{1}{L}\sin(\omega t)\) 2. Find the integrating factor: \(e^{\frac{R}{L}t}\) 3. Multiply the entire equation by the integrating factor: \(e^{\frac{R}{L}t}\frac{di}{dt} + e^{\frac{R}{L}t}\frac{R}{L}i = \frac{1}{L}\sin(\omega t)e^{\frac{R}{L}t}\) 4. Solve the modified equation by integration: \(\frac{d}{dt}\left(e^{\frac{R}{L}t}i\right) = \frac{1}{L}\sin(\omega t)e^{\frac{R}{L}t}\) 5. Solve for the general solution: \(i(t) = e^{-\frac{R}{L}t}I\) where I is an integral involving sin(ωt) and the exponential function.

Step-by-step solution

Rewrite the ODE in standard form

The given ODE can be written in its standard form: \(\frac{di}{dt} + \frac{R}{L}i = \frac{1}{L}\sin(\omega t)\) Here, we have divided both sides of the equation by L. Our goal is to solve for i(t). ##Step 2: Find the integrating factor##

Find the integrating factor

In general, the integrating factor for a first order linear ODE of the form: \(\frac{dy}{dx} + P(x)y = Q(x)\) is: \(e^{\int P(x) dx}\) Here, we have: P(x) = \(\frac{R}{L}\) (constant) Integrating P(x) with respect to t, we get: \(\int \frac{R}{L} dt\) = \(\frac{R}{L}t\) Hence, the integrating factor (IF) will be: \(e^{\frac{R}{L}t}\) ##Step 3: Multiply the entire equation by the integrating factor##

Multiply the entire equation by the integrating factor

Now, multiply the entire equation by the integrating factor: \(e^{\frac{R}{L}t}\frac{di}{dt} + e^{\frac{R}{L}t}\frac{R}{L}i = \frac{1}{L}\sin(\omega t)e^{\frac{R}{L}t}\) ##Step 4: Solve the modified equation by integration##

Solve the modified equation by integration

The left side of the modified equation can be written as the derivative of i(t) multiplied by the integrating factor: \(\frac{d}{dt}\left(e^{\frac{R}{L}t}i\right) = \frac{1}{L}\sin(\omega t)e^{\frac{R}{L}t}\) Now, integrate both sides with respect to t: \(\int \frac{d}{dt}\left(e^{\frac{R}{L}t}i\right) dt= \int \frac{1}{L}\sin(\omega t)e^{\frac{R}{L}t}dt\) Let this integral on the right-hand side be I: \(e^{\frac{R}{L}t}i = I\) ##Step 5: Solve for the general solution##

Solve for the general solution

To find i(t), we simply divide both sides of the equation obtained in Step 4 by the integrating factor: \(i(t) = e^{-\frac{R}{L}t}I\) Here, I is an integral involving sin(ωt) and the exponential function, which may or may not provide a known function. In either case, this is the general solution of the ODE when multiplied with the integrating factor.

Key concepts

Understanding Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving derivatives of a function and the function itself. They are called 'ordinary' to distinguish them from 'partial' differential equations, which involve partial derivatives of a function with respect to multiple variables.

ODEs appear in various scientific and engineering disciplines to model dynamic systems, whose state changes with one independent variable, usually time. Solving an ODE means finding a function that satisfies the equation. For instance, consider the equation given in the exercise, where the variable i(t) represents some physical quantity that varies with time, such as electrical current:

\[\begin{equation}L\frac{di}{dt} + Ri = \text{sin}(\text{omega} t),\end{equation}\]
The equation comprises a function i(t), its derivative \(\frac{di}{dt}\), and a sinusoidal source term. It describes a dynamic situation where the current i(t) changes over time due to both a resistive element R and an inductive element L in an electric circuit when subject to a sinusoidal driving force.

Exponential Functions in Differential Equations

The exponential function, denoted by \(e^x\), where e is the base of the natural logarithm, is vitally important in mathematics, especially when dealing with differential equations. Its unique properties make it an invaluable tool for solving equations where the rate of change of a quantity is proportional to the quantity itself.

In the context of the integrating factor method, the exponential function is utilized to facilitate the solution of the first-order linear ODE. An integrating factor is a function, typically an exponential, that transforms a non-exactable differential equation into an exactable one. In other words, it helps turn the original ODE into a form where the left-hand side becomes the derivative of a product of two functions. Thus, the solution can be obtained by direct integration.

For the given exercise, the integrating factor is: \(e^{\frac{R}{L}t}\). Multiplying the original ODE through by this integrating factor ensures the left side of the equation conforms to the derivative of the product of i(t) and the integrating factor itself, allowing the equation to be integrated effectively.

Integration by Parts in Solving Differential Equations

Integration by parts is a technique based on the product rule for differentiation and is used to integrate products of functions. The equation that defines integration by parts is:

\[\begin{equation}\int u dv = uv - \int v du\end{equation}\]
In the final step of the solving process, as seen in the exercise, the integration by parts might be required to evaluate the integral involving the product of sinusoidal and exponential functions:

\[\begin{equation}I = \int \frac{1}{L}\text{sin}(\text{omega} t)e^{\frac{R}{L}t} dt\end{equation}\]
Here, the standard choice for using integration by parts would be to set one part of the integrand to u (usually the part that becomes simpler upon differentiation), and the other to dv (usually the part that becomes simpler upon integration). Depending on the integrand's complexity, this method might simplify the solution process or even yield a closed-form expression for the integral.