Problem 32 - Thomas Calculus

Question

a. Show that the solution of the equation $$\frac{d i}{d t}+\frac{R}{L} i=\frac{V}{L}$$ is $$i=\frac{V}{R}+C e^{-(R / L) x}$$ b. Then use the initial condition \(i(0)=0\) to determine the value of \(C .\) This will complete the derivation of Equation \((7)\) . c. Show that \(i=V / R\) is a solution of Equation \((6)\) and that \(i=C e^{-(R / L) x}\) satisfies the equation $$\frac{d i}{d t}+\frac{R}{L} i=0$$

Step-by-Step Solution

Answer

The value of \( C \) is \(-\frac{V}{R}\).

Step 1: Solve the Homogeneous Equation

Start by considering the homogeneous part of the equation: \( \frac{d i}{d t} + \frac{R}{L} i = 0 \). This is a linear equation with constant coefficients. The solution is of the form \( i_h = C e^{ax} \), where \( a = -\frac{R}{L} \). Thus, the homogeneous solution is \( i_h = C e^{-\frac{R}{L} x} \).

Step 2: Find Particular Solution

To find a particular solution \( i_p \), assume \( i_p = \frac{V}{R} \). Differentiating gives \( \frac{d i_p}{d t} = 0 \). Substitute \( i_p \) into the original equation: \( 0 + \frac{R}{L} \frac{V}{R} = \frac{V}{L} \). The equation holds true, confirming \( i_p = \frac{V}{R} \) is a particular solution.

Step 3: General Solution

The general solution of the differential equation is the sum of the homogeneous and particular solutions: \[ i = i_h + i_p = C e^{-\frac{R}{L} x} + \frac{V}{R} \].

Step 4: Apply Initial Condition

Use the initial condition \( i(0) = 0 \) to find \( C \). Substitute \( x = 0 \) into the general solution: \( 0 = Ce^{0} + \frac{V}{R} \), which simplifies to \( 0 = C + \frac{V}{R} \). Thus, \( C = -\frac{V}{R} \).

Step 5: Substitute Value of C

Substitute \( C = -\frac{V}{R} \) back into the general solution: \[ i = \frac{V}{R} - \frac{V}{R} e^{-\frac{R}{L} x} \].

Step 6: Verify Particular Solution 1

To verify \( i = \frac{V}{R} \), differentiate: \( \frac{d i}{d t} = 0 \). Check: \( 0 + \frac{R}{L} \frac{V}{R} = \frac{V}{L} \), which holds, confirming it satisfies the original equation.

Step 7: Verify Homogeneous Solution

For \( i = C e^{-\frac{R}{L} x} \), differentiate: \( \frac{d i}{d t} = -\frac{R}{L} C e^{-\frac{R}{L} x} \). Substitute into the equation: \( -\frac{R}{L} C e^{-\frac{R}{L} x} + \frac{R}{L} C e^{-\frac{R}{L} x} = 0 \). Both terms cancel, verifying that it satisfies the homogeneous equation.

Key Concepts

Linear Differential Equations

In mathematics, linear differential equations play a pivotal role in describing systems with known rates of change. A differential equation is called "linear" when the dependent variable and all its derivatives appear to the power of one, without any products of the dependent variable with itself or with its derivatives. For example, the linear differential equation \( \frac{d i}{d t} + \frac{R}{L} i = \frac{V}{L} \) describes the relationship between the electric current \( i \), resistance \( R \), inductance \( L \), and voltage \( V \).

• The general form of a first-order linear differential equation is \( \frac{dy}{dt} + P(t)y = Q(t) \).
• In our equation, \( P(t) = \frac{R}{L} \) and \( Q(t) = \frac{V}{L} \).
• Methods for finding solutions often involve finding the homogeneous solution and particular solution.

Particular Solution

Identifying a particular solution is a crucial step in solving linear differential equations. This solution accounts for the non-homogeneous part of the equation. For our equation \( \frac{d i}{d t} + \frac{R}{L} i = \frac{V}{L} \), the particular solution is found by assuming a constant solution where the derivative \( \frac{d i_p}{d t} = 0 \).

• Assume \( i_p = \frac{V}{R} \), leading to \( \frac{d i_p}{d t} = 0 \).
• Substitute \( i_p \) back into the equation to verify: \( 0 + \frac{R}{L} \cdot \frac{V}{R} = \frac{V}{L} \).
• This value of \( i_p \) satisfies the original differential equation, confirming that \( i_p = \frac{V}{R} \) is indeed a particular solution.

The particular solution is key because it directly represents a specific solution without involving arbitrary constants, reflecting the system's steady state.

Homogeneous Solution

The homogeneous solution deals with the differential equation when no external forces or inputs are considered, meaning the right-hand side is zero. For the equation \( \frac{d i}{d t} + \frac{R}{L} i = 0 \), the solution reflects the natural behavior of the system.

• To find the homogeneous solution, consider the equation without the external input: \( \frac{d i}{d t} + \frac{R}{L} i = 0 \).
• Assume the solution is of the form \( i_h = Ce^{ax} \), where \( a = -\frac{R}{L} \).
• Thus, \( i_h = C e^{-(R/L)x} \), which describes exponentially decaying behavior.

The homogeneous solution captures the transient behavior of the system and depends on the initial conditions, eventually giving way to the steady-state behavior represented by the particular solution.

Initial Conditions

Initial conditions are critical in determining the specific values of constants in solutions to differential equations. They ensure that the solution fits the real-world scenario from a specific starting point. Given the general solution \( i = \frac{V}{R} + C e^{-(R/L) x} \), if \( i(0) = 0 \), then substituting these conditions allows solving for \( C \).

• Substitute the initial condition into the general solution: \( 0 = \frac{V}{R} + C e^{0} \).
• This simplifies to \( 0 = \frac{V}{R} + C \), thus \( C = -\frac{V}{R} \).
• The initial conditions effectively "pin down" the value of \( C \), giving the complete solution adapted to the particular situation.

By applying the initial condition, we transition from a generic form to one that meets the specified requirements, ensuring accuracy and relevance in real-life applications.