Problem 19 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(11-20,\) solve the initial value problem explicitly. \(\frac{d v}{d t}=4 \sec t \tan t+e^{t}+6 t\) and \(v=5\) when \(t=0\)

Step-by-Step Solution

Answer

Therefore, the solution to the initial value problem is \(v(t)=4\sec t + e^{t} + 3t^{2}\).

Step 1: Integration

Integrate the first order non-homogeneous ordinary differential equation by isolating the \(\frac{dv}{dt}\) on one side and integrating the other side with respect to \(t\) to get \(v(t)\). \nThus, \(\int dv = \int (4\sec t\tan t+e^t +6t)dt)\) leading to \(v(t) = 4\int(\sec t\tan t) dt + \int(e^{t})dt + \int(6t)dt + C\), where \(C\) is the constant of integration.

Step 2: Evaluate Integrals

Evaluate each integral separately. All of these are standard integrals: \(\int\sec t\tan t dt=\sec t\), \(\int e^t dt=e^t\) and \(\int 6t dt=3t^2\). Thus, \(v(t)=4\sec t + e^{t} + 3t^{2} + C\).

Step 3: Apply Initial Condition

Use the initial condition \(v=5\) when \(t=0\) to solve for the constant \(C\). Substituting these values into the equation we get: \(5 = 4(1) + 1 + 0 + C\). Solving for \(C\) yields \(C = 0\).

Step 4: Final Solution

Substitute \(C=0\) back into the equation from step 2 to get the final solution to the initial value problem, which is \(v(t)=4\sec t + e^{t} + 3t^{2}\).

Key Concepts

Ordinary Differential Equation

An ordinary differential equation (ODE) involves a function and its derivatives. It is defined in terms of a single independent variable, often representing time. The equation given, \( \frac{dv}{dt}=4 \sec t \tan t+e^{t}+6t \), is an ODE because it contains the derivative \( \frac{dv}{dt} \). Here, \( v \) is the dependent variable depending on \( t \), the independent variable.

ODEs can describe a wide variety of phenomena in physics, engineering, and other fields by showing relationships between variables and their rates of change.

• First-order ODE: when it contains the first derivative of the unknown function.
• Non-homogeneous ODE: if there is a non-zero function on the right side of the equation, as in our case.

Solving an ODE often requires techniques such as separation of variables, integrating factors, or in our case, direct integration.

Integration

Integration is a mathematical process that helps find a function when its derivative is known. In solving ODEs, integration is often used to solve for the function in terms of the independent variable.

In this problem, we integrate both sides of the equation \( \int dv = \int (4 \sec t \tan t+e^{t}+6t)dt \). Integration tells us how to accumulate quantities over time, which, in our case, results in the function \( v(t) \).

• Separating integrals: Each term on the right can be integrated separately to simplify calculations.
• Recognizing Standard Integrals: E.g., \( \int \sec t \tan t dt = \sec t \), uses known integral results directly.

Integration takes advantage of the fundamental theorem of calculus, connecting antiderivatives and definite integrals, to find solutions.

Constant of Integration

The constant of integration, represented by \( C \), emerges whenever we integrate a function. This constant accounts for all the possible values that a function could take after integration, given that the derivative of a constant is zero.

For the equation \( v(t) = 4 \sec t + e^{t} + 3t^{2} + C \), the constant of integration \( C \) represents any constant value that could adjust the function \( v(t) \) to match specific conditions.

• When solving an initial value problem, \( C \) is initially undetermined until additional conditions (like the initial condition) are applied.
• It helps in finding the unique solution for the specific conditions of a problem.

In real-world problems, the constant ensures the solution fits the scenario being modeled.

Initial Condition

When solving differential equations, the initial condition is a critical piece of information provided to help find a particular solution. It reduces the set of possible solutions to just one by determining the value of the constant of integration.

In this exercise, the initial condition \( v = 5 \) when \( t = 0 \) means, at time \( t=0 \), the function \( v(t) \) equals 5. This information is used to solve for \( C \).

• By substituting \( t = 0 \) and \( v = 5 \) into the equation, we solve \( 5 = 4(1) + 1 + 0 + C \) to find \( C = 0 \).
• Initial conditions tailor general solutions of differential equations to meet specific requirements or observations.

These conditions reflect starting values and are essential in many scientific and engineering applications.