Problem 17 - Calculus: Early Transcendentals

Question

Consider the differential equation \(a y^{\prime \prime}+b y^{\prime}=0,\) with a and \(b\) both non-zero. Find the general solution by the method of this section. Now let \(g=y^{\prime} ;\) the equation may be written as \(a g^{\prime}+b g=0,\) a first order linear homogeneous equation. Solve this for \(g,\) then use the relationship \(g=y^{\prime}\) to find \(y\).

Step-by-Step Solution

Answer

The general solution is \( y = C \frac{a}{b} e^{\frac{b}{a} t} + D \).

Step 1: Rewrite the Equation

The given differential equation is \( a y'' + b y' = 0 \). We need to solve this by rewriting it using a substitution. Let \( g = y' \). Thus the equation becomes \( a g' + b g = 0 \).

Step 2: Identify the Type of Equation

The equation \( a g' + b g = 0 \) is a first-order linear homogeneous differential equation. We can solve it using the technique for first-order linear homogeneous equations.

Step 3: Solve for g

The standard form of a first-order linear homogeneous differential equation is \( g' + \frac{b}{a} g = 0 \). The solution can be found using the integrating factor method. The integrating factor \( \mu(t) \) is \( e^{\int -\frac{b}{a} dt} = e^{-\frac{b}{a} t} \). Multiply the entire equation by this integrating factor to get \[ e^{-\frac{b}{a} t} g' + \frac{b}{a} e^{-\frac{b}{a} t} g = 0. \]

Step 4: Integrate to Find g

Notice that the left-hand side is the derivative of \( e^{-\frac{b}{a} t} g \): \[ \frac{d}{dt} \left( e^{-\frac{b}{a} t} g \right) = 0. \] Integrate both sides w.r.t \( t \), \( e^{-\frac{b}{a} t} g = C \), where \( C \) is a constant. Therefore, \( g = C e^{\frac{b}{a} t} \).

Step 5: Solve for y

Recall that \( g = y' \), so \( y' = C e^{\frac{b}{a} t} \). Integrate \( y' \) to find \( y \): \[ y = \int C e^{\frac{b}{a} t} \, dt = C \frac{a}{b} e^{\frac{b}{a} t} + D, \] where \( D \) is another constant of integration.

Key Concepts

First-Order Linear Homogeneous Equation

A first-order linear homogeneous differential equation is a foundational concept in differential equations. These types of equations take the form \(a_1(t)y' + a_0(t)y = 0\), where the solution is focused on the derivative \(y'\) and the function \(y\) itself. In the equation \(a y'' + b y' = 0\), we performed a substitution using \(g = y'\), resulting in the form \(a g' + b g = 0\). This is indeed a first-order linear homogeneous equation because it relates \(g'\), the derivative of \(g\), to \(g\) itself without any constant terms. Homogeneous means that every term involves \(y\) or its derivatives. Such equations demonstrate the principle of superposition, where any constant multiplied by a solution is also a solution.

• Linear means each term is to the first degree.
• Homogeneous means there are no constant terms added.
• Only the function and its derivatives are present.

Understanding this form is crucial because it simplifies recognizing and solving the equation using standard techniques, such as the integrating factor method. The solution of a homogeneous linear equation often leads to exponential functions, which are naturally suited to matching the properties of the derivatives.

Integrating Factor Method

The integrating factor method is a powerful tool for solving first-order linear differential equations. It helps convert an easily solvable differential equation into one involving separable variables. In our context, the original differential equation, after substitution, was \(a g' + b g = 0\). By dividing through by \(a\), it takes the standard form \(g' + \frac{b}{a}g = 0\). To solve for \(g\), we use an integrating factor \(\mu(t)\), which is determined by the expression \(e^{\int -\frac{b}{a} dt} = e^{-\frac{b}{a} t}\). This factor transforms the equation into one where the left side becomes the derivative of a product.

• The integrating factor simplifies the equation, allowing for direct integration.
• This technique works broadly for linear first-order equations, beyond just homogeneous ones.
• After applying \(\mu(t)\), the equation becomes easier to integrate directly.

Applying the integrating factor method changes the differential equation into one where the solution is straightforwardly obtained by integration. This technique is crucial for turning seemingly complex equations into manageable expressions involving simple functions.

General Solution

The goal of solving a differential equation is often to find its general solution, an expression involving all possible particular solutions. After altering the original problem into \(g = y'\) and solving for \(g\), we found that \(g = C e^{\frac{b}{a} t}\). Here, \(C\) is an arbitrary constant. To find the general solution for \(y\), we integrate \(g = y'\) with respect to \(t\). This led us to the equation:\[y = \int C e^{\frac{b}{a} t} dt = C \frac{a}{b} e^{\frac{b}{a} t} + D\]Here, \(D\) represents another constant of integration.

• The general solution signifies an entire family of solutions due to its arbitrary constants.
• These constants allow the solutions to adapt to initial or boundary conditions.
• The form \(C \frac{a}{b} e^{\frac{b}{a} t} + D\) highlights both the homogeneous response and integrating constants.

General solutions are essential because they provide the most comprehensive form, anchoring the scope to include any particular solution with the correct choice of constants. This approach ensures every potential behavior of the system described by the differential equation is covered.