Problem 5 - Calculus for Biology and Medicine

Question

Solve each pure-time differential equation. \(\frac{d x}{d t}=\frac{1}{1-t}\), where \(x(0)=2\)

Step-by-Step Solution

Answer

The solution is \( x(t) = -\ln|1-t| + 2 \).

Step 1: Separate Variables

The given differential equation is \( \frac{dx}{dt} = \frac{1}{1-t} \). This equation is already in a form that separates the variables as it aligns with integrating both sides with respect to \( t \).

Step 2: Integrate Both Sides

Integrate both sides of the equation with respect to \( t \): \[ x(t) = \int \frac{1}{1-t} dt \] The integral of \( \frac{1}{1-t} \) is \( -\ln|1-t| + C \), where \( C \) is the integration constant.

Step 3: Solve for the Integration Constant

Use the initial condition \( x(0) = 2 \) to solve for \( C \): \[ 2 = -\ln|1-0| + C \] Since \( \ln(1) = 0 \), it implies \( C = 2 \).

Step 4: Write the Particular Solution

Substitute \( C = 2 \) back into the integrated equation to find the particular solution: \[ x(t) = -\ln|1-t| + 2 \]

Key Concepts

Initial Conditions

When solving differential equations, initial conditions play a pivotal role. They provide the specific values needed to determine the exact solution for a problem. An initial condition gives the value of the solution, or its derivatives, at a particular point. In the context of our exercise, the initial condition is given as \( x(0) = 2 \).
This means when time \( t \) is zero, the function \( x(t) \) is equal to 2.

Initial conditions are critical because differential equations often have infinite solutions.

• These conditions help us narrow down to one specific solution.
• Without them, determining the unique, relevant solution for a particular context is not possible.

For our equation, the initial condition \( x(0) = 2 \) allows us to solve for the integration constant \( C \) after integrating. It ensures that our solution not only satisfies the equation but also adheres to the specified circumstances.

Integration

Integration is a mathematical technique that combines or adds up small elements to construct a whole. In calculus, it's used to find areas, volumes, central points, and other cumulative attributes.

In our exercise, integration comes into play in step 2, where the equation \( x(t) = \int \frac{1}{1-t} \, dt \) is derived. To solve this, you need to find the antiderivative of \( \frac{1}{1-t} \), which is \( -\ln|1-t| + C \).

This integration yields a general solution that includes an arbitrary integration constant \( C \):

• \( C \) represents an and justifiable number.
• The presence of \( C \) expresses the infinite family of solutions for an indefinite integral.

The process of integration is crucial in solving differential equations as it helps transition from the derivative form back to a function form. It’s important to recognize how the limits shift in the process of definite and indefinite integration.

Separation of Variables

Separation of variables is a common technique used to solve differential equations. It's particularly effective for equations where the variables can be separated on different sides of the equation.

In our problem, the differential equation \( \frac{dx}{dt} = \frac{1}{1-t} \) is already ideal for this method. This technique involves:

• Isolating the variables on opposite sides of the equation.
• It transforms the equation into a form where each side contains only one variable.

By doing this, each side can be individually integrated. It facilitates easier integration and simplifies the problem.

In practice, separation of variables transforms the differential equation into a form that can be more straightforwardly handled and integrated. It's a technique based on rearranging terms so every variable stands alone with its differential counterpart.
When done correctly, it makes solving differential equations an achievable task, often leading to solutions that can be addressed with clear initial conditions.