Problem 9 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The general solution is \( x(t) = e^{-\frac{t^2}{2} + t} \left( \int e^{\frac{t^2}{2} - t}(t-1) \, dt + C \right) \).

Step 1: Identify the Equation Type

The given differential equation \( \frac{dx}{dt} + (t-1)x = t-1 \) is a first-order linear differential equation of the form \( \frac{dx}{dt} + P(t)x = Q(t) \) where \( P(t) = t-1 \) and \( Q(t) = t-1 \).

Step 2: Find the Integrating Factor

To solve using an integrating factor, calculate \( \mu(t) = e^{\int P(t) \, dt} = e^{\int (t-1) \, dt} \). This integral results in \( \mu(t) = e^{\frac{t^2}{2} - t} \).

Step 3: Multiply by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \mu(t) \) to obtain \( e^{\frac{t^2}{2} - t} \frac{dx}{dt} + e^{\frac{t^2}{2} - t}(t-1)x = e^{\frac{t^2}{2} - t}(t-1) \).

Step 4: Rewrite as a Derivative

Recognize that the left-hand side is the derivative of \( e^{\frac{t^2}{2} - t} x \), so rewrite the equation as \( \frac{d}{dt}\left(e^{\frac{t^2}{2} - t}x\right) = e^{\frac{t^2}{2} - t}(t-1) \).

Step 5: Integrate Both Sides

Integrate both sides with respect to \( t \), giving \( e^{\frac{t^2}{2} - t}x = \int e^{\frac{t^2}{2} - t}(t-1) \, dt + C \), where \( C \) is the constant of integration.

Step 6: Solve for the General Solution

To solve for \( x(t) \), divide both sides by the integrating factor \( e^{\frac{t^2}{2} - t} \), yielding \( x(t) = e^{-\frac{t^2}{2} + t} \left( \int e^{\frac{t^2}{2} - t}(t-1) \, dt + C \right) \).

Key Concepts

Integrating Factor

When solving first-order linear differential equations, the integrating factor is a crucial tool. It helps transform the differential equation into a form that is more easily solvable. The integrating factor, denoted by \(\mu(t)\), is derived from the function \(P(t)\) in the standard form \(\frac{dx}{dt} + P(t)x = Q(t)\).To find the integrating factor, you calculate:

• \(\mu(t) = e^{\int P(t) \, dt}\)

In our example, since \(P(t) = t-1\), the integrating factor is:

• \(\mu(t) = e^{\int (t-1) \, dt} = e^{\frac{t^2}{2} - t}\)

The integrating factor essentially acts as a multiplier for the entire differential equation, setting the stage for combining terms into a complete derivative. This way, we can integrate directly, making it much easier to find the solution.

First-Order Linear Differential Equation

First-order linear differential equations involve derivatives of a function and the function itself but only to the first power. The standard form of a first-order linear differential equation is:

• \(\frac{dx}{dt} + P(t)x = Q(t)\)

Here, \(P(t)\) and \(Q(t)\) are functions of \(t\). The equation we have, \(\frac{dx}{dt} + (t-1)x = t-1\), fits this format perfectly, with:

• \(P(t) = t-1\)
• \(Q(t) = t-1\)

Understanding the structure allows us to apply systematic approaches, such as finding an integrating factor, to solve the equation efficiently. Recognizing this type of equation is the first step in applying the method of integrating factors for an elegant solution.

General Solution

The general solution of a differential equation encompasses all possible solutions, incorporating an arbitrary constant. This constant represents the freedom in selecting any one of the possible solutions based on specific initial conditions.In the process of solving our first-order linear differential equation, the goal was to get it into the form that allowed integration. After rewriting the equation with the integrating factor, we arrived at:

• \(e^{\frac{t^2}{2} - t}x = \int e^{\frac{t^2}{2} - t}(t-1) \, dt + C\)

By dividing both sides by the integrating factor, we solve for the general solution \(x(t)\). Including the constant \(C\) allows the general solution to adapt to different initial conditions, making it a powerful expression to describe the behavior of the system over time.

Method of Integration

Once a differential equation is prepped with an integrating factor, the next step is applying the method of integration. Here, integration is used to find the antiderivative, essentially "summing up" small contributions of change to find the total effect.After manipulating the equation with an integrating factor, it simplifies to a form where the left side is the derivative of a product of functions. Our rewritten equation became:

• \(\frac{d}{dt}\left(e^{\frac{t^2}{2} - t}x\right) = e^{\frac{t^2}{2} - t}(t-1)\)

Integrating both sides with respect to \(t\) gives:

• \(e^{\frac{t^2}{2} - t}x = \int e^{\frac{t^2}{2} - t}(t-1) \, dt + C\)

This step effectively solves the differential equation, providing us with a direct pathway to arrive at the general solution. It exemplifies the elegance of integration in simplifying and solving complex mathematical problems.