Problem 7 - Calculus: Early Transcendentals

Question

Find the general solution of the equation. $$t y^{\prime}-2 y=1 / t, t>0$$

Step-by-Step Solution

Answer

The general solution is \( y = -\frac{1}{3t} + Ct^2 \).

Step 1: Rewrite the equation in standard form

Start with the given equation \( t y' - 2y = \frac{1}{t} \) and rewrite it in the standard form for a first-order linear differential equation. Divide each term by \( t \) to isolate \( y' \):\[ y' - \frac{2}{t}y = \frac{1}{t^2} \].

Step 2: Identify the integrating factor

The standard form of the differential equation is \( y' + P(t)y = Q(t) \). Here, \( P(t) = -\frac{2}{t} \). The integrating factor \( \mu(t) \) is determined by:\[ \mu(t) = e^{\int P(t) \, dt} = e^{-2 \int \frac{1}{t} \, dt} = e^{-2 \ln t} = t^{-2} \].

Step 3: Multiply through by the integrating factor

Multiply the entire differential equation by the integrating factor \( t^{-2} \):\[ t^{-2} y' - t^{-2} \cdot \frac{2}{t} y = t^{-2} \cdot \frac{1}{t^2} \],which simplifies to:\[ t^{-2} y' - \frac{2}{t^3} y = \frac{1}{t^4} \].

Step 4: Write the left side as a derivative

Recognize the left-hand side as the derivative of the product \( (t^{-2}y) \' \):\[ \frac{d}{dt}(t^{-2}y) = \frac{1}{t^4} \].

Step 5: Integrate both sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(t^{-2}y) \, dt = \int \frac{1}{t^4} \, dt \].The left side simplifies by cancellation. The right side integrates to:\[ t^{-2}y = -\frac{1}{3t^3} + C \],where \( C \) is the constant of integration.

Step 6: Solve for y

Solve for \( y \) by multiplying through by \( t^2 \):\[ y = -\frac{1}{3t} + Ct^2 \].This expresses \( y \) in terms of \( t \) and the constant \( C \).

Key Concepts

Linear Differential Equations

Linear differential equations are a special class of differential equations characterized by their linearity in terms of the unknown function and its derivatives. This means that the unknown function, let's say \( y \), and its derivative \( y' \), appear to the power of one and are not multiplied together or by themselves.
Linear differential equations can usually be expressed in the form \( a(t) y' + b(t) y = c(t) \), where \( a(t) \), \( b(t) \), and \( c(t) \) are given functions of the variable \( t \).
Some key characteristics include:

• Superposition Principle: Solutions of a linear differential equation adhere to the principle of superposition. This means that the sum of two solutions is also a solution.
• Homogeneity: If \( c(t) = 0 \), the equation is homogeneous; otherwise, it is non-homogeneous.

Understanding these properties allows us to apply systematic methods like the integrating factor to find solutions.

Integrating Factor

An integrating factor is a mathematical tool used to solve linear first-order differential equations. It transforms a non-exact differential equation into an exact one, making it easier to solve. The integrating factor is often denoted by \( \mu(t) \).
For a standard linear differential equation of the form \( y' + P(t)y = Q(t) \), the integrating factor is computed as:
\[ \mu(t) = e^{\int P(t) \, dt} \]
Once the integrating factor is determined, the differential equation is multiplied throughout by this factor. This process simplifies the equation into a form where the left side can be rewritten as the derivative of a product.

• This step is crucial because it allows us to apply integration to both sides easily, ultimately finding the solution.

With the integrating factor applied, the differential equation becomes much more tractable, particularly in our context where we tackle first-order linear differential equations.

First-Order Differential Equations

First-order differential equations involve the first derivative of the unknown function and perhaps the function itself. These equations are fundamental in studying various physical phenomena due to their straightforward nature.
The standard form of a first-order linear differential equation is \( y' + P(t)y = Q(t) \). In this configuration:

• \( y' \) represents the first derivative of the unknown function \( y \).
• \( P(t) \) and \( Q(t) \) are functions of \( t \), potentially constants or more complex functions.

The simplicity of first-order equations means they often have exact or neat solutions. Utilizing tools like the integrating factor further simplifies the process of finding these solutions.
In many practical situations, first-order differential equations model exponential growth or decay, electrical circuits, or thermal dynamics.

General Solution

The general solution of a differential equation comprises all possible specific solutions. It represents a family of solutions dependent on one or more arbitrary constants. For a first-order linear differential equation, the general solution might look something like \( y(t) = \, function \, of \, t \, and \, C \), where \( C \) is the constant.
The significance of finding a general solution is that it encompasses every specific solution that satisfies the differential equation under given initial conditions.
To determine a general solution:

• Use an integrating factor to rewrite the differential equation in an exact form.
• Integrate both sides to find \( y(t) \) as a function of \( t \) and \( C \).

This process is demonstrated in scenarios like the exercise provided, where after applying the integrating factor and integrating, we solve for \( y \) in terms of \( t \) and \( C \), capturing the complete set of potential solutions.