Problem 55 - Calculus for Biology and Medicine

Question

In Problems 55-60 you will need to solve differential equations by separation of variables. In these problems it will not always be possible to solve explicitly for \(y\) in terms of \(t ;\) instead your solution may take the form of an implicit function relating the two variables. $$ \frac{d y}{d t}=\frac{y^{2}+y}{t-1} \text { where } y(0)=1 $$

Step-by-Step Solution

Answer

The implicit solution is \( \frac{|y|}{|y+1|} = \frac{|t-1|}{2} \).

Step 1: Separate Variables

Start by rearranging the differential equation to separate the variables, so all terms involving \( y \) are on one side and all terms involving \( t \) are on the other.Given: \( \frac{dy}{dt} = \frac{y^2 + y}{t-1} \).Separate variables to obtain:\[ \frac{dy}{y^2 + y} = \frac{dt}{t-1}. \]

Step 2: Integrate Both Sides

Integrate both sides of the equation:For the left side: \( \int \frac{1}{y^2 + y} \, dy \).For the right side: \( \int \frac{1}{t-1} \, dt \).Right side: \[ \int \frac{1}{t-1} \, dt = \ln |t-1| + C_1 \].Left side requires partial fraction decomposition: \(\frac{1}{y^2 + y} = \frac{1}{y(y+1)} = \frac{A}{y} + \frac{B}{y+1} \).Solving gives \( A = 1 \) and \( B = -1 \). Integrate:\[ \int \left( \frac{1}{y} - \frac{1}{y+1} \right) \, dy = \ln |y| - \ln |y+1| + C_2. \]

Step 3: Combine Results

Combine the integrated results:\[ \ln |y| - \ln |y+1| = \ln |t-1| + C. \]Where \( C = C_1 - C_2 \).

Step 4: Use Initial Condition

Apply the initial condition \( y(0) = 1 \) to find \( C \).Substitute \( y = 1 \) and \( t = 0 \) into the equation:\[ \ln |1| - \ln |1+1| = \ln |0-1| + C. \]Simplifies to \(- \ln 2 = \ln 1 + C \) hence \( C = - \ln 2 \).

Step 5: Solve Implicitly

Substitute \( C \) back into the equation:\[ \ln |y| - \ln |y+1| = \ln |t-1| - \ln 2. \]Rearrange to express implicitly:\[ \frac{|y|}{|y+1|} = \frac{|t-1|}{2}. \]

Key Concepts

Separation of Variables

Differential equations often require techniques to simplify and solve them, and separation of variables is one of the most straightforward and powerful methods. In this technique, the goal is to rearrange the equation so that all terms involving one variable are on one side and those involving another variable are on the other side. This allows each variable by itself to be integrated separately.

In the given exercise, the differential equation is \( \frac{dy}{dt} = \frac{y^2 + y}{t-1} \). We start by separating the variables. The terms involving \( y \) are moved to one side, and those involving \( t \) to the other, leading to:

• \( \frac{dy}{y^2 + y} = \frac{dt}{t-1} \)

Once separated, each side of the equation can be integrated separately. The beauty of this technique lies in its ability to transform a differential equation into integrable forms, making the problem more approachable.

Implicit Functions

After separation and integration, solutions to some differential equations cannot always be explicitly written as \( y = f(t) \). Instead, they might express a relationship between \( y \) and \( t \) without solving for one variable in terms of the other. These types of solutions are called implicit functions.

In our problem, after integration and applying the initial condition, we reached:

• \( \ln |y| - \ln |y+1| = \ln |t-1| - \ln 2 \)

This expression doesn't solve for \( y \) explicitly as a function of \( t \). Rather, it indicates a relationship between \( y \) and \( t \). Such implicit solutions are often encountered in scenarios where equations are complex or when variables influence each other in non-linear ways. Implicit forms capture the essence of the solution even without direct separation of the dependent and independent variables.

Initial Conditions

Initial conditions offer specific information that helps narrow down possible solutions of differential equations. They are essential for finding a unique solution from a general solution set. An initial condition specifies the value of the unknown function at a particular point, for example when \( t = 0 \), \( y = 1 \).

In this example, the initial condition is \( y(0) = 1 \). By substituting \( y = 1 \) and \( t = 0 \) into our implicit equation:

• \( \ln |1| - \ln |1+1| = \ln |0-1| + C \)

Upon simplifying, it helps find the constant \( C \). Initial conditions transform the general solution into one that fits a specific context or scenario, providing the particular solution which is essential in many real-world applications. Without initial conditions, differential equations would have a myriad of possible solutions.