Question

Using software, obtain a symbolic solution of \(L \frac{\mathrm{d} i}{\mathrm{~d} t}+R i=10\left(1-\mathrm{e}^{-0.1 t}\right)\) when \(i(0)=i_{0}\)

Short answer

Question: Solve the following ordinary differential equation (ODE) for the current "i" as a function of time "t", considering the initial condition i(0) = i0: $$L \frac{di}{dt}+R i=10(1-e^{-0.1 t})$$ Answer: The solution for the current "i" as a function of time "t" is given by: $$i(t) = \frac{10}{L}h(t)e^{-\frac{R}{L}t} + (i_0 - \frac{10}{L}h(0))e^{-\frac{R}{L}t}$$ where \(h(t)\) is the integral of \((1-e^{-0.1 t})e^{\frac{R}{L}t}\) with respect to "t".

Step-by-step solution

Identify differential equation

The differential equation given is: $$L \frac{di}{dt}+R i=10(1-e^{-0.1 t})$$ with the initial condition i(0) = i0.

Rewrite the differential equation

Rewrite the given differential equation as a first-order linear ODE in standard form: $$\frac{di}{dt}+\frac{R}{L}i = \frac{10}{L}(1-e^{-0.1 t})$$

Calculate the integrating factor

Find the integrating factor: $$\mu(t) = e^{\int \frac{R}{L}dt} = e^{\frac{R}{L}t}$$

Multiply the ODE with the integrating factor

Multiply both sides of the given ODE by the integrating factor: $$e^{\frac{R}{L}t}(\frac{di}{dt}+\frac{R}{L}i) = \frac{10}{L}e^{\frac{R}{L}t}(1-e^{-0.1 t})$$

Integration

Now integrate both sides of the equation with respect to time "t": $$\int\left[e^{\frac{R}{L}t}(\frac{di}{dt}+\frac{R}{L}i)\right]dt = \int\left[\frac{10}{L}e^{\frac{R}{L}t}(1-e^{-0.1 t})\right]dt$$

Simplification

The left side of the equation is simply the derivative of the product i(t) and the integrating factor. For the right side of the equation, we need to use integration by parts or other integration techniques. After integrating, the equation becomes: $$i(t)e^{\frac{R}{L}t}= \frac{10}{L}\int e^{\frac{R}{L}t}(1-e^{-0.1 t})dt + C$$ Now, divide both sides of the equation by \(e^{\frac{R}{L}t}\) to get i(t): $$i(t) = \frac{10}{L}h(t)e^{-\frac{R}{L}t} + Ce^{-\frac{R}{L}t}$$ where \(h(t)\) is the integral of \((1-e^{-0.1 t})e^{\frac{R}{L}t}\) with respect to "t".

Find the constant C

Apply the initial condition i(0) = i0: $$i0 = \frac{10}{L}h(0) + C$$ Solve for C, and substitute it back into the equation in Step 6.

Final Answer

After substituting the constant C, we get the expression for the current "i" as a function of time "t": $$i(t) = \frac{10}{L}h(t)e^{-\frac{R}{L}t} + (i_0 - \frac{10}{L}h(0))e^{-\frac{R}{L}t}$$

Key concepts

First-Order Linear ODE

A differential equation is a mathematical equation that relates some function with its derivatives. In the context of the exercise given, we are concerned with a first-order linear ordinary differential equation (ODE). This type of equation is first-order because it involves only the first derivative of the function (in this case, \(\frac{di}{dt}\)) and it is linear because the function and its derivative appear to the first power and are not multiplied together.

Linear ODEs are foundational in the study of differential equations, as they can model a wide range of physical phenomena, including electrical circuits, which is the context of our exercise. A typical first-order linear ODE is expressed in the format: \[\frac{dx}{dt} + p(t)x = q(t)\], where \(x\) is the unknown function of time \(t\), and \(p(t)\) and \(q(t)\) are known functions. The solution to such equations can be specific (solved for particular initial conditions) or general (solving for a function in terms of the variable).

Understanding these equations is crucial because, for many systems, the first order is sufficient to provide an accurate model for analysis. In our exercise, the current over time \(i(t)\) in an electrical circuit is determined by solving a first-order linear ODE.

Integrating Factor Method

Determining the solution to a linear ODE can be facilitated by using the integrating factor method. This robust technique involves multiplying the entire equation by a specially chosen function called the integrating factor which simplifies the equation into an integrable form.

The integrating factor, usually denoted by \(\mu(t)\), is determined by the formula \[\mu(t) = e^{\int p(t)dt}\], where \(p(t)\) is the coefficient of \(x\) from the standard form of the first-order linear ODE. By multiplying the original ODE by \(\mu(t)\), we convert the left side of the equation into the derivative of the product of \(\mu(t)\) and \(x(t)\), which allows for simple integration.

The power of the integrating factor method lies in its ability to streamline complex problems, making them more accessible and solvable. As demonstrated in the exercise, once the integrating factor is found, the solution process involves standard algebraic manipulations and basic integration techniques. The method can turn what initially appears to be a daunting differential equation into one which can be deftly integrated to yield a solution.

Initial Conditions

To find a specific solution to a differential equation, one must take into account the initial conditions of the system being modeled. Initial conditions describe the state of a system at the beginning of the observation period, usually when \(t = 0\). They are instrumental for solving differential equations uniquely.

For instance, in the exercise, we are given an initial condition \(i(0) = i_0\), which means that the value of the current \(i(t)\) at time \(t = 0\) is \(i_0\). This piece of information is crucial for determining the constant of integration \(C\) after finding the general solution of the differential equation. In other words, initial conditions help 'lock in' the solution to a particular context or scenario.

Without initial conditions, we can only provide a general solution, which includes an arbitrary constant. However, with initial conditions, we can solve for that constant, making our solution specific to the given situation. This is what ultimately allows us to model real-world systems accurately and predict their future behavior based on their current state.