Problem 31 - Advanced Engineering Mathematics

Question

For \(0 \leq t \leq 20\) the differential equation is \(20 d i / d t+2 i=120 .\) An integrating factor is \(e^{t / 10},\) so \((d / d t)\left[e^{t / 10} i\right]=\) \(6 e^{t / 10}\) and \(i=60+c_{1} e^{-t / 10} .\) If \(i(0)=0\) then \(c_{1}=-60\) and \(i=60-60 e^{-t / 10} .\) For \(t>20\) the differential equation is \(20 d i / d t+2 i=0\) and \(i=c_{2} e^{-t / 10} .\) At \(t=20\) we want \(c_{2} e^{-2}=60-60 e^{-2}\) so that \(c_{2}=60\left(e^{2}-1\right)\). Thus $$i(t)=\left\\{\begin{array}{ll} 60-60 e^{-t / 10}, & 0 \leq t \leq 20 \\ 60\left(e^{2}-1\right) e^{-t / 10}, & t>20 \end{array}\right.$$

Step-by-Step Solution

Answer

\(i(t)=\begin{cases} 60 - 60e^{-t/10}, & 0 \leq t \leq 20 \\ 60(e^2 - 1)e^{-t/10}, & t > 20 \end{cases}\)

Step 1: Identify the differential equation for each interval

For the interval \(0 \leq t \leq 20\), the differential equation is \(20 \frac{di}{dt} + 2i = 120\). For "\(t > 20\), the differential equation becomes \(20 \frac{di}{dt} + 2i = 0\)."

Step 2: Find the integrating factor for the first interval

For the equation \(20 \frac{di}{dt} + 2i = 120\), divide through by 20 to get \(\frac{di}{dt} + \frac{1}{10}i = 6\). The integrating factor is \(e^{\frac{t}{10}}\).

Step 3: Solve using the integrating factor

Multiply through by the integrating factor: \(e^{\frac{t}{10}} \times \frac{di}{dt} + e^{\frac{t}{10}} \times \frac{1}{10}i = 6e^{\frac{t}{10}}\). This simplifies to \(\frac{d}{dt}(e^{\frac{t}{10}}i) = 6e^{\frac{t}{10}}\).

Step 4: Integrate to find a general solution for the first interval

Integrate both sides with respect to \(t\): \ \(e^{\frac{t}{10}}i = \int 6e^{\frac{t}{10}} dt = 60e^{\frac{t}{10}} + C_1\). Solving for \(i\) gives \(i = 60 + C_1e^{-\frac{t}{10}}\).

Step 5: Apply initial condition for the first interval

Given that \(i(0) = 0\), substitute into the equation \(0 = 60 + C_1 e^0\), which gives \(C_1 = -60\). Hence, \(i = 60 - 60e^{-\frac{t}{10}}\) for \(0 \leq t \leq 20\).

Step 6: Solve the differential equation for the second interval

For \(t > 20\), \(20 \frac{di}{dt} + 2i = 0\). Dividing through by 20: \(\frac{di}{dt} + \frac{1}{10}i = 0\). The solution is \(i = C_2 e^{-\frac{t}{10}}\).

Step 7: Match solutions at \(t = 20\)

At \(t = 20\), the first interval's solution gives \(i = 60 - 60e^{-2}\). For continuity, set this equal to \(C_2 e^{-2}\), yielding \(C_2 = 60(e^2 - 1)\).

Step 8: Write the piecewise solution

Combine both solutions to form a piecewise function: \ \(i(t) = \begin{cases} 60 - 60e^{-\frac{t}{10}}, & 0 \leq t \leq 20 \ 60(e^2 - 1)e^{-\frac{t}{10}}, & t > 20 \end{cases}\).

Key Concepts

Integrating Factor

When dealing with linear first-order differential equations, the integrating factor is an incredibly useful method. Essentially, it transforms a non-exact equation into an exact one, making it easier to solve. The integrating factor, often denoted as \(e^{\int P(t) \, dt}\), is a function that, once multiplied to all terms of the differential equation, allows the left side of the equation to become the derivative of a product of functions.

• For example, consider \( \frac{di}{dt} + \frac{1}{10}i = 6 \). Here, the integrating factor is \( e^{\frac{t}{10}} \).
• This transforms the equation into \( \frac{d}{dt}(e^{\frac{t}{10}}i) = 6e^{\frac{t}{10}} \), which can be integrated more straightforwardly than the original form.

By converting the problem into one involving exact differentials, the integrating factor method makes solving these equations systematically manageable, bridging the known functions of \(i\) and its derivatives.

Initial Condition

When solving differential equations, particularly when finding specific solutions rather than general ones, it is crucial to apply initial conditions. Initial conditions are values that specify the state of the system at a particular point, often at the start of observation.

• In our context, given \(i(0) = 0\) as an initial condition, we substitute this into the general solution of the first interval.
• This condition allows us to solve for any arbitrary constants present in our solution, ensuring the solution fits the specific case.
• For example, this helps us determine that \(C_1 = -60\) by setting \(0 = 60 + C_1\).

Initial conditions are vital in providing context-specific solutions and ensuring that the calculated function accurately reflects the system's behavior from the designated start time.

Piecewise Function

A piecewise function is a function defined by multiple sub-functions, each applicable to a certain interval of the independent variable's domain. It is particularly useful when different behaviors or rules apply to different parts of a problem.

• For instance, the solution to our differential equation involves two intervals: \(0 \leq t \leq 20\) and \(t > 20\).
• The solution for each interval reflects the different conditions governing the system. Between \(0\) and \(20\), the system's behavior is described by \(i = 60 - 60e^{-\frac{t}{10}}\).
• After \(20\), the behavior shifts, leading to the piece \(i = 60(e^2 - 1)e^{-\frac{t}{10}}\).
• By using a piecewise function, we can concisely express these distinct behaviors within a single formula.

Piecewise functions allow for concise expression of variable-dependent behavior over specified intervals, making them ideal for real-world processes that change at certain points in time.