Problem 4 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The general solution is \( y(t) = t^2 - 2t + 2 + Ce^{-t} \), where \( C \) is a constant.

Step 1: Identify the type of differential equation

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

Step 2: Find the integrating factor

The integrating factor \( \mu(t) \) is given by \( \mu(t) = e^{\int P(t) \, dt} \). In this case, \( \mu(t) = e^{\int 1 \, dt} = e^t \).

Step 3: Multiply through by the integrating factor

Multiply every term in the differential equation by the integrating factor \( e^t \): \[ e^t \frac{dy}{dt} + e^t y = t^2 e^t \].

Step 4: Rewrite as a derivative

The left side of the equation \( e^t \frac{dy}{dt} + e^t y \) can be written as the derivative of a product: \[ \frac{d}{dt}(e^t y) = t^2 e^t \].

Step 5: Integrate both sides

To solve for \( y \), integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(e^t y) \, dt = \int t^2 e^t \, dt \].The left side simplifies to \( e^t y \). Integrate the right side using integration by parts, setting \( u = t^2 \) and \( dv = e^t dt \).

Step 6: Solve the integration by parts

From integration by parts, \( du = 2t \, dt \) and \( v = e^t \), so:\[ \int t^2 e^t \, dt = t^2 e^t - \int 2t e^t \, dt. \].Repeat integration by parts on \( \int 2t e^t \, dt \) with \( u=t \).

Step 7: Complete the solution for \( y \)

After completing integration by parts twice, we get:\[ \int t^2 e^t \, dt = t^2 e^t - 2(te^t - e^t) + C = t^2 e^t - 2te^t + 2e^t + C. \]Equating the simplified integrated expression: \[ e^t y = t^2 e^t - 2te^t + 2e^t + C. \].Solve for \( y \): \[ y = t^2 - 2t + 2 + Ce^{-t}. \]

Step 8: State the general solution

The general solution of the differential equation is \[ y(t) = t^2 - 2t + 2 + Ce^{-t}, \]where \( C \) is an arbitrary constant.

Key Concepts

Integrating Factors

The concept of integrating factors is crucial when solving linear first-order differential equations. An integrating factor is a function that we multiply through the entire differential equation to make it possible to rewrite the equation in a convenient form.
Doing this allows the equation to be transformed into one that can be directly integrated with respect to the independent variable.
For the differential equation \[ \frac{dy}{dt} + P(t)y = Q(t), \] the integrating factor, \( \mu(t) \), is given by: \[ \mu(t) = e^{\int P(t) \, dt}. \]

• Multiply every term of the differential equation by \( \mu(t) \).
• This transforms the left-hand side into a derivative of a product.

Once the derivative form is achieved, the equation can be integrated easily, solving for the unknown function, \( y \). Integrating factors are a clever way to deal with equations that otherwise seem resistant to straightforward integration.

Linear Differential Equation

A linear differential equation involves derivatives of a function and is linear in the variables involved. These equations have the form: \[ \frac{dy}{dt} + P(t)y = Q(t), \]where:

• \( \frac{dy}{dt} \): The first derivative of \( y \) with respect to \( t \).
• \( P(t) \) and \( Q(t) \): Functions of \( t \), specifying the coefficients.

Linear differential equations are important in many applications.
They often represent physical laws and processes, which makes understanding and solving them significant in fields like physics and engineering.
For example, the cooling of an object or the charging of a capacitor can be described by such equations.
The solution process typically involves using techniques like integrating factors to simplify and solve them.

Integration by Parts

Integration by parts is a method used to integrate products of functions. This technique is vital when dealing with integration especially in calculus.
In essence, it relies on the rule of differentiation stating that: \[ \int u \, dv = uv - \int v \, du. \]

• Choose \( u \) and \( dv \) from the original integral.
• Find \( du \) by differentiating \( u \), and find \( v \) by integrating \( dv \).
• Substitute into the integration by parts formula.

In the given problem, integrating by parts was essential in exploring the relationship between \( t^2 \) and \( e^t \).
This method can simplify complex integrations into a more manageable form, facilitating the completion of the problem.
While the integration can become recursive, repeating the method only brings us closer to the solution.

First-Order Differential Equations

First-order differential equations are equations involving the first derivative of the unknown function.
These equations are fundamental as they model a variety of natural phenomena, including population growth, radioactive decay, and heat flow.These equations can generally be written as:\[ \frac{dy}{dt} = f(t, y). \]They can be linear, as the example discussed, or non-linear depending on the nature of \( f(t, y) \).
In our case, the equation is linear, making it suitable for the integrating factor method.
Understanding first-order differential equations is key to exploring more advanced differential topics.
Solutions of first-order equations often involve integration or applying special methods like separation of variables or, as seen here, an integrating factor that simplifies the integration.