Problem 14 - 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 . ~. $$ \frac{d y}{d t}=\frac{y}{t+1} $$

Step-by-Step Solution

Answer

Use separation of variables to solve: the solution is \( y = C(t+1) \).

Step 1: Identify the Type of Differential Equation

First, we need to determine if the given differential equation \( \frac{dy}{dt} = \frac{y}{t+1} \) can be solved using separation of variables or the method of integrating factors. The equation is in the form \( \frac{dy}{dt} = f(y,t) \), where \( f(y,t) = \frac{y}{t+1} \). It is suitable for separation of variables because \( f(y,t) \) can be expressed as the product of a function of \( y \) alone and a function of \( t \) alone: i.e., \( f(y,t) = y \cdot \frac{1}{t+1} \).

Step 2: Rewrite in Separable Form

Rewrite the equation in such a way to separate the variables. We have:\[ \frac{dy}{y} = \frac{1}{t+1} \, dt \]

Step 3: Integrate Both Sides

Now integrate both sides of the equation:\[ \int \frac{dy}{y} = \int \frac{1}{t+1} \, dt \]This yields:\[ \ln |y| = \ln |t+1| + C \]where \( C \) is the integration constant.

Step 4: Solve for y

Exponentiate both sides to solve for \( y \):\[ |y| = e^{\ln |t+1| + C} = |t+1| \cdot e^C \]Set \( e^C = C' \), where \( C' \) is a new constant. Thus, \( y = C' (t+1) \). \( C' \) can be positive or negative to account for the absolute value.

Step 5: State the General Solution

The general solution of the differential equation is \[ y = C(t+1) \]where \( C \) is an arbitrary constant.

Key Concepts

Separation of Variables

Separation of variables is a delightful technique when tackling differential equations. Imagine you are trying to separate peanut butter from jelly in a sandwich – not exactly easy, but not rocket science either! Here, we do something similar, but with mathematical terms. The goal is to separate the variable terms that involve the dependent variable (such as \(y\)) from those involving the independent variable (such as \(t\)).

In our example, the differential equation is \(\frac{dy}{dt} = \frac{y}{t+1}\). Those terms can be thought of as 'peanut butter' and 'jelly'. The method involves rearranging the equation to form two sides, one purely \(y\) and the other \(t\), hence why it is called 'separation of variables'.

• Step 1: Identify the form \(f(y,t) = g(y)h(t)\).
• Step 2: Manipulate to \(\frac{1}{g(y)} \frac{dy}{dt} = h(t)\).
• Step 3: Integrate both sides separately.
• Step 4: Solve for the dependent variable \(y\).

With separation of variables, it's almost like having each variable find its own path, leading to a much simpler integration and overall solution.

Integrating Factors

Integrating factors come to the rescue when dealing with linear first-order differential equations where separation does not work as well. While separation of variables is about simplification, integrating factors aim to transform the equation into something that is easily recognizable and solvable – kind of like turning a Rubik's Cube until one side matches at least!

Here’s how an integrating factor steps in:

• Step 1: Check if the differential equation is linear. It must be in form \(dy/dt + P(t)y = Q(t)\).
• Step 2: Compute the integrating factor, typically \(\mu(t) = e^{\int P(t) dt}\).
• Step 3: Multiply the entire differential equation by \(\mu(t)\), turning it into a much simpler form.
• Step 4: Integrate the new and tidy form easily.

It's much like having a secret tool in your toolbox: sometimes, equations need that special touch to be nicely answerable! No integrating factor was necessary in our problem since separation worked beautifully, but it’s good to know when this method can be the knight in shining armor.

Calculus

Calculus is the backbone of solving differential equations – it provides the tools needed for integration and differentiation. Solving differential equations often feels like playing a detective game, finding clues to solve for unknowns. With integration, we add up tiny pieces to find the whole picture, whereas differentiation breaks everything into tiny pieces.

In the problem \(\frac{dy}{dt} = \frac{y}{t+1}\), calculus allows us to integrate both sides of the separated equation to find the function \(y\):

• The left side integrates the function with respect to \(y\).
• The right side integrates the function with respect to \(t\).
• Results lead to a logarithmic function \(\ln |y| = \ln |t+1| + C\).
• The exponential operation recovers \(y\) by 'undoing' the logarithm.

Understanding calculus is like possessing the keys to a locked treasure – once you've learned to integrate and differentiate, you're well-equipped to solve many differential equations and see the beauty of underlying patterns and trends.