Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Solve the following differential equations. Give a general solution and then a particular solution corresponding to the initial condition given. \(\begin{array}{ll}\text { (a) } \frac{d y}{d t}=3 y & \text { initial condition: } y(0)=5\end{array}\) (b) \(\frac{d y}{d x}=-0.01 y \quad\) initial condition: \(y(2)=1\) (c) \(\frac{d w}{d s}=w\) initial condition: \(w(0)=\pi\)

Step-by-Step Solution

Answer

The particular solutions for the exercises are: (a) \(y(t)=5e^{3t}\), (b) \(y(x)=1.0202e^{-0.01x}\), and (c) \(w(s)=\pi e^{s}\).

Step 1: Solve the general solution for (a)

To solve \( \frac{dy}{dt} = 3y \), separate the variables by dividing both sides by y and multiplying both sides by dt. It gives us \( \frac{1}{y} dy = 3 dt \). Integrate both sides to get \(\int \frac{1}{y} dy = \int 3dt\). Solving the integrals gives \(\ln|y| = 3t + C\), where C is the constant of integration. Solving for y gives the general solution: \(y(t) = Ce^{3t}\) where C is determined by the initial condition.

Step 2: Find particular solution for (a)

Use the initial condition y(0)=5 in the general solution. Substitute 0 for t and 5 for y(t) in \(y(t) = Ce^{3t}\), yielding \(5 = Ce^{0}\), which simplifies to \(5 = C\). Thus, the particular solution is \(y(t) = 5e^{3t}\).

Step 3: Solve the general solution for (b)

The differential equation for (b) is \( \frac{dy}{dx}=-0.01y \). Solve it using similar steps as in step 1 to get the general solution, \(y(x) = Ce^{-0.01x}\).

Step 4: Find particular solution for (b)

Use the initial condition y(2)=1 to find the value of C. Substituting 2 for x and 1 for y(x) in \(y(x) = Ce^{-0.01x}\) would give us \(1 = Ce^{-0.02}\) , and solving for C gives \(C \approx 1.0202\). Thus, the particular solution is \(y(x) = 1.0202e^{-0.01x}\).

Step 5: Solve the general solution for (c)

The differential equation for (c) is \(\frac{dw}{ds}=w\). Following the same method as before, the general solution will be \(w(s) = Ce^{s}\).

Step 6: Find particular solution for (c)

Using the initial condition w(0)= π to find the value of C. Substituting 0 for s and π for w(s) in \(w(s) = Ce^{s}\) gives \(π = Ce^{0}\) which simplifies to \(π = C\). Therefore, the particular solution is \(w(s) = \pi e^{s}\).

Key Concepts

General Solution of Differential Equation

When dealing with differential equations, the general solution represents a family of solutions containing one or more arbitrary constants. It embodies all possible solutions to the differential equation. For instance, in the exercise provided, the differential equation \( \frac{dy}{dt} = 3y \) has a general solution of the form \( y(t) = Ce^{3t} \) for the variable \( y \) as a function of \( t \) where \( C \) is a constant. This constant \( C \) will not have a value until we apply an initial condition or additional information.

Understanding the general solution is crucial because it shows the behavior of the dynamic system described by the differential equation without being restricted to a particular scenario. The process typically involves separation of variables and integration to find an expression that includes an arbitrary constant representing the range of solutions.

Particular Solution of Differential Equation

A particular solution to a differential equation, on the other hand, is obtained when specific initial conditions are applied to the general solution. This solution satisfies not only the differential equation but also the initial values given in the problem. For example, in the first problem of the provided exercise \( y(0) = 5 \) the particular solution is found by determining the constant \( C \). Plugging \( t = 0 \) and \( y(0) = 5 \) into the general solution \( y(t) = Ce^{3t} \) yields \( 5 = Ce^{0} \) which simplifies to \( 5 = C \) thus giving us the particular solution; \( y(t) = 5e^{3t} \).

The particular solution is unique and shows how the system behaves under the specific initial conditions provided, offering a concrete example of the system's dynamics at a certain point in time.

Separation of Variables

The separation of variables method is a common technique used to solve a particular type of differential equations. To separate the variables, one must manipulate the equation so that each variable and its differential appear on opposite sides of the equation. For the differential equation \( \frac{dy}{dt} = 3y \) from our exercise, the procedure is to divide by \( y \) so that \( \frac{1}{y}dy = 3dt \).

After the variables have been separated, the next step involves the integration of functions to find the general solution. It's essential to perform the separation accurately, as incorrect manipulation can lead to unsolvable integrals or incorrect solutions. One must be precise in handling the differentials and integrating functions to ensure a correct and useful general solution.

Integration of Functions

The process of integration of functions is an essential step in solving differential equations, particularly after separating variables. Integration transforms the differential expressions into algebraic ones. In our exercise, after separating variables, we integrate both sides of the equation \( \frac{1}{y}dy = 3dt \) to obtain \(\int \frac{1}{y} dy = \int 3dt\), which results in \(\ln|y| = 3t + C\), where \(C\) is the constant of integration.

The integral of a function reveals the accumulated value over a range, and in the context of differential equations, it contributes to finding the general solution that satisfies the equation. One must master a variety of integration techniques such as substitution, integration by parts, and partial fractions to tackle different differential equations effectively.