Problem 1

Question

Find the general solution of each equation in the following exercises. $$y^{\prime}+5 y=0$$

Step-by-Step Solution

Verified

Answer

The general solution is $ y = Ce^{-5x} $.

1Step 1: Identify Equation Type

The given equation is a first-order linear homogeneous differential equation of the form $ y' + ay = 0 $, where $ a = 5 $.

2Step 2: Write the General Form Solution

For a first-order linear homogeneous differential equation $ y' + ay = 0 $, the general solution is given by $ y = Ce^{-ax} $, where $ C $ is an arbitrary constant.

3Step 3: Substitute and Simplify

Substitute $ a = 5 $ into the general form $ y = Ce^{-ax} $. This gives $ y = Ce^{-5x} $. The solution is simplified to its final form.

Key Concepts

Linear Homogeneous Differential EquationGeneral SolutionArbitrary Constant

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is a specific type of differential equation, characterized by its linearity and homogeneity. It is typically expressed in standard form as follows:

The equation involves derivatives, such as $ y' $, of a function $ y $.
Linearity means that the equation involves no powers or products of the function or its derivatives beyond the first degree.
Homogeneity indicates that every term in the equation involves the dependent variable or its derivatives.

For example, the differential equation $ y' + 5y = 0 $ exemplifies this form. Here, $ a = 5 $, representing the coefficient that multiplies the function $ y $. Linear homogeneous equations are important because they are simpler forms that help us understand more complex dynamics.

General Solution

The general solution of a linear homogeneous differential equation is a broad answer that includes all particular solutions of the differential equation. This generality stems from the way it incorporates an arbitrary constant, enabling it to reflect an entire family of solutions, rather than a single, isolated result.To derive the general solution for an equation like $ y' + ay = 0 $, with

$ a $ as a constant coefficient, we often use the exponential function due to its relationship with derivatives.
The resulting general solution takes the form $ y = Ce^{-ax} $, where $ C $ is an arbitrary constant.

By this method, the equation $ y' + 5y = 0 $ turns into $ y = Ce^{-5x} $. This general solution can be adapted for any initial condition by selecting an appropriate value of $ C $.

Arbitrary Constant

The arbitrary constant $ C $ in the solution of a differential equation is a vital element because it introduces flexibility into the solution. This constant is not specified within the equation itself and must be determined through additional conditions, typically known as initial or boundary conditions.By incorporating $ C $, the general solution $ y = Ce^{-ax} $ embodies many specific solutions, each unique to particular initial values or constraints.

In problems without provided initial conditions, $ C $ remains as an undetermined parameter, signifying the solution's general nature.
Given conditions like $ y(x_0) = y_0 $, $ C $ can be computed to yield a single, particular solution fitting the context.

Hence, this arbitrary constant is what allows differential equations to be universally applicable across different scenarios.

Problem 1

Other exercises in this chapter

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In the following exercises, compute the Euler approximations for the initial value problem for $0 \leq t \leq 1$ and $\Delta t=0.2 .$ If you have access to

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Problem 2

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