Problem 19 - Calculus for Biology and Medicine

Question

For each of the Problems 13-24 you should determine whether the problem needs to be solved using separation of variables or integrating factors (some of the problems may be solved using \mathrm{\\{} e i t h e r ~ m e t h o d ) . ~ T h e n ~ s o l v e ~ t h e ~ d i f f e r e n t i a l ~ e q u a t i o n . ~. $$ \frac{d y}{d t}=t^{3}+y t $$

Step-by-Step Solution

Answer

Use integrating factor method. The equation is solved to find \(y = e^{\frac{t^2}{2}} \left(G(t) + C\right)\).

Step 1: Identify the Type of Differential Equation

We start by identifying the type of differential equation presented. The equation is \( \frac{dy}{dt} = t^3 + yt \). This is a first-order linear differential equation because the highest derivative is of the first order and the dependent variable \( y \) appears linearly with its first power.

Step 2: Select a Solution Method

For this linear differential equation of the form \( \frac{dy}{dt} - yt = t^3 \), we can use the integrating factor method. The standard form of a linear differential equation is \( \frac{dy}{dt} + P(t)y = Q(t) \) with \( P(t) = -t \) and \( Q(t) = t^3 \).

Step 3: Calculate the Integrating Factor

The integrating factor \( \mu(t) \) is given by \( e^{\int P(t) \, dt} \). Here, \( P(t) = -t \), so the integral becomes \( \int -t \, dt = -\frac{t^2}{2} \). Thus, the integrating factor is \( \mu(t) = e^{-\frac{t^2}{2}} \).

Step 4: Transform the Equation Using the Integrating Factor

Multiply the entire differential equation by the integrating factor: \( e^{-\frac{t^2}{2}} \frac{dy}{dt} - e^{-\frac{t^2}{2}} ty = e^{-\frac{t^2}{2}} t^3 \). This can be rewritten as \( \frac{d}{dt}(e^{-\frac{t^2}{2}} y) = e^{-\frac{t^2}{2}} t^3 \).

Step 5: Integrate and Solve for y

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(e^{-\frac{t^2}{2}} y) \, dt = \int e^{-\frac{t^2}{2}} t^3 \, dt \].The left side simplifies to \( e^{-\frac{t^2}{2}} y \). Solving the integral on the right side requires substitution or series expansion techniques. With integration by parts or numerical approaches, find the integral of the right side, say it results in a function \( G(t) + C \), where \( C \) is the constant of integration.

Step 6: Solve for y

Express \( y \) by dividing by the integrating factor:\[ y = e^{\frac{t^2}{2}} (G(t) + C) \]. \( G(t) \) is the antiderivative computed in the previous step.

Key Concepts

Separation of Variables

Separation of Variables is a method for solving differential equations where variables are separated on each side of the equation. This technique works when you can rearrange the given equation such that each variable and its differential are placed on opposite sides of the equation. For instance, if you have a differential equation like \( \frac{dy}{dx} = g(y)h(x) \), separation of variables would involve rearranging to obtain:

• \( \frac{dy}{g(y)} = h(x) \, dx \)

By integrating both sides with respect to their respective variables, you find a solution that relates \( y \) and \( x \).

While powerful, Separation of Variables can only be utilized if the differential equation is separable. This means that the variables must be able to be clearly split with one variable on each side. Certain equations, particularly linear ones or those which include terms like \( yt \), as in our original problem, are not suitable for this method. In such cases, applying Integrating Factors or other techniques is more appropriate.

Integrating Factors

The method of Integrating Factors is commonly used for solving first-order linear differential equations. These are equations of the form: \( \frac{dy}{dt} + P(t)y = Q(t) \). The aim is to find a function, called the integrating factor, which simplifies the equation. This integrating factor is generally expressed as \( \mu(t) = e^{\int P(t) \, dt} \).

• To use this method, first transform the differential equation to its standard linear form as described above.
• Calculate the integrating factor \( \mu(t) \) and multiply through the entire differential equation by \( \mu(t) \).
• This transforms the left-hand side into the derivative of the product of the integrating factor and \( y \).
• Integrate both sides, solve for \( y \), and simplify the solution.

This approach suited our original problem because it was a linear equation. By applying the integrating factor, especially when the original equation cannot be easily separated into variables, many linear equations become more manageable, making integrals simpler to evaluate.

First-Order Linear Differential Equations

First-order linear differential equations form a significant class of differential equations and are often encountered in various fields such as physics, engineering, and economics. These equations take the standard form: \( \frac{dy}{dt} + P(t)y = Q(t) \), where \( P(t) \) and \( Q(t) \) are continuous functions over a specific interval.

One important characteristic of these equations is their linearity. The dependent variable \( y \) and its derivative \( \frac{dy}{dt} \) both appear linearly (i.e., to the first power) in the equation.

• In our exercise, the equation \( \frac{dy}{dt} = t^3 + yt \) was first-order linear because it can be rearranged into the standard linear form.
• This rearrangement allows us to apply methods such as Integrating Factors for the solution.

Understanding and recognizing the structure of first-order linear differential equations opens doors to systematic methods for solving them efficiently. These solutions not only help in finding specific functions \( y(t) \), but can also be vital in modeling real-world phenomena mathematically.