Problem 3 - Applied Calculus

Question

Find solutions to the differential equations in subject to the given initial condition. $$\frac{d y}{d x}=-0.14 y, \quad y=5.6 \text { when } x=0$$

Step-by-Step Solution

Answer

The solution is \( y = 5.6 e^{-0.14x} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{d y}{d x} = -0.14 y \). This is a simple first-order linear differential equation with constant coefficients, which is also a separable differential equation.

Step 2: Separate the Variables

Rearrange the equation to separate the variables \( y \) and \( x \):\[\frac{d y}{y} = -0.14 \, d x\]The terms involving \( y \) are on one side, and the terms involving \( x \) are on the other side.

Step 3: Integrate Both Sides

Integrate both sides of the equation:\[\int \frac{d y}{y} = \int -0.14 \, d x\]This gives:\[\ln|y| = -0.14x + C\]where \( C \) is the constant of integration.

Step 4: Solve for y

To solve for \( y \), exponentiate both sides to remove the natural logarithm:\[|y| = e^{-0.14x + C} = e^C \cdot e^{-0.14x}\]Since \( e^C \) is just a constant, we can denote it as \( C_1 \), so:\[y = C_1 e^{-0.14x}\]

Step 5: Apply the Initial Condition

Use the initial condition \( y = 5.6 \) when \( x = 0 \) to find \( C_1 \):\[5.6 = C_1 e^{0}\]Thus, \( C_1 = 5.6 \).

Step 6: Write the Particular Solution

Substitute \( C_1 \) back into the equation for \( y \):\[y = 5.6 e^{-0.14x}\]This is the particular solution to the differential equation subject to the given initial condition.

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a type of differential equation where the highest derivative is the first derivative. It takes a specific form, which is:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

The equation in our exercise, \( \frac{dy}{dx} = -0.14y \), fits this description but is even simpler due to its constant coefficient.
This structure is beneficial because it often allows for straightforward solution techniques, such as separation of variables or integrating factors.
Linear differential equations are widely used because they model many real-world phenomena, especially where rates of change are involved. Understanding their structure helps in solving them more efficiently. It sets the foundation for tackling more complex differential equations later.

Separable Differential Equations

Separable differential equations enable us to isolate the variables completely, making them easier to solve. Such equations have the form:

• \( g(y) dy = h(x) dx \)

This allows us to integrate each side independently to find a solution.
In the exercise, the given differential equation \( \frac{dy}{dx} = -0.14y \) is rearranged to \( \frac{dy}{y} = -0.14 \, dx \), effectively separating the variables.
Solving separable equations involves integrating both sides. This technique is especially handy when dealing with equations describing phenomena like exponential growth or decay.
Understanding separable differential equations builds strong foundational skills, as it simplifies complex real-world problems into more manageable mathematical equations.

Initial Value Problem

An initial value problem (IVP) provides not only a differential equation but also a specific condition that the solution must satisfy. This condition, known as the initial value, is given at a particular point.
For the example in the exercise, we have the initial value \( y = 5.6 \) when \( x = 0 \). This allows us to determine constant values in the solution after integrating.
The process of solving an IVP involves:

• Solving the differential equation generally.
• Using the initial condition to find specific constants.
• Substituting back to get a particular solution.

IVPs are critical in applications where knowing the starting point is crucial, such as predicting future behavior based on initial conditions in physics or engineering problems.

Constant Coefficients

Constant coefficients in differential equations are coefficients that do not depend on the variable. They greatly simplify solving differential equations.
In our exercise, the coefficient \(-0.14\) is constant, which allows for straightforward techniques like separation of variables or the characteristic equation method.
With constant coefficients, differential equations often model scenarios with uniform properties, such as steady-state heat flow or consistent rate of decay.
Understanding how to handle constant coefficients enables easier solutions and is an essential skill when dealing with linear differential equations, making it a fundamental topic in differential equations.