Problem 1 - Calculus for Biology and Medicine

Question

Find the general solution of the differential equations in Problems 1-12 using the method of integrating factors: $$ \frac{d y}{d t}+\frac{y}{t}=\frac{1}{t^{2}} $$

Step-by-Step Solution

Answer

The general solution is \( y = \frac{\ln|t|}{t} + \frac{C}{t} \).

Step 1: Identify the Standard Form

The differential equation is given by \( \frac{d y}{d t} + \frac{y}{t} = \frac{1}{t^2} \). This is a first-order linear differential equation of the form \( \frac{d y}{d t} + P(t) y = Q(t) \). Here, \( P(t) = \frac{1}{t} \) and \( Q(t) = \frac{1}{t^2} \).

Step 2: Compute the Integrating Factor

The integrating factor \( \mu(t) \) is calculated using the formula \( \mu(t) = e^{\int P(t) \, dt} \). In this case, \( \mu(t) = e^{\int \frac{1}{t} \, dt} = e^{\ln |t|} = |t| \). Since \( t > 0 \) for our solution context, it simplifies to \( t \).

Step 3: Multiply Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( t \):\[ t \frac{dy}{dt} + y = \frac{1}{t} \].This transforms the equation into one that can be written in the form of an exact derivative on the left-hand side.

Step 4: Express as Exact Derivative

Notice that the left side can be written as a derivative: \( \frac{d}{dt}(ty) = t \frac{dy}{dt} + y \). Thus, we have:\[ \frac{d}{dt} (ty) = \frac{1}{t} \].

Step 5: Integrate Both Sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt} (ty) \, dt = \int \frac{1}{t} \, dt \].The left side integrates to \( ty \), giving us:\[ ty = \ln|t| + C \], where \( C \) is the constant of integration.

Step 6: Solve for y

Solve for \( y \) by dividing both sides by \( t \): \[ y = \frac{\ln|t|}{t} + \frac{C}{t} \]. This is the general solution of the differential equation.

Key Concepts

Integrating Factor

An integrating factor is a very useful mathematical tool that simplifies the process of solving first-order linear differential equations. It helps in transforming a non-exact equation into an exact one, making it easier to solve. The key idea is to multiply the whole differential equation by a strategically chosen function, called the integrating factor, to simplify the equation.

For a first-order linear differential equation of the form \[ \frac{dy}{dt} + P(t)y = Q(t) \], the integrating factor \( \mu(t) \) is given by:

• \( \mu(t) = e^{\int P(t) \, dt} \)

In the original exercise, the differential equation is \( \frac{dy}{dt} + \frac{y}{t} = \frac{1}{t^2} \). Here, \( P(t) = \frac{1}{t} \), making the integrating factor \( e^{\int \frac{1}{t} \, dt} = t \).

Multiplying the entire differential equation by this integrating factor \( t \) yields an equation in a form that's simpler to solve. The integrating factor process reduces the problem complexity and makes subsequent steps much more straightforward.

First-Order Linear Differential Equation

A first-order linear differential equation is a differential equation where the highest derivative is of the first order, and it can be written in the standard form:

• \( \frac{dy}{dt} + P(t) y = Q(t) \)

Here,

• \( \frac{dy}{dt} \) is the derivative of \( y \) with respect to \( t \).
• \( P(t) \) and \( Q(t) \) are known functions of \( t \).

In our particular problem, we have \( P(t) = \frac{1}{t} \) and \( Q(t) = \frac{1}{t^2} \). This form allows us to apply specific solution techniques like using an integrating factor to simplify and solve the equation.

Such equations are fundamental in modeling many real-world phenomena, including population growth models, cooling of objects, and electrical circuits, among many others. Recognizing the form allows us to apply these techniques effectively.

General Solution

The general solution of a differential equation is a solution that contains all possible particular solutions. In other words, it represents the family of all solutions to the differential equation, typically containing one or more arbitrary constants.

In our example of the differential equation \( \frac{dy}{dt} + \frac{y}{t} = \frac{1}{t^2} \), after converting, simplifying, and integrating through the integrating factor technique, we obtained:

• \( y = \frac{\ln|t|}{t} + \frac{C}{t} \)

Here, \( C \) is the arbitrary constant of integration, representing the plethora of solutions depending on initial conditions or constraints.

Understanding the general solution is crucial because it highlights the potential outcomes under varied circumstances. Adjusting \( C \) allows fitting the solution to specific boundary conditions or initial values, thus tailoring it to unique situations. This flexibility is key in practical applications.