Problem 4 - Calculus for Biology and Medicine

Question

Solve each pure-time differential equation. \(\frac{d y}{d x}=\frac{1}{1+x^{2}}\), where \(y_{0}=1\) when \(x_{0}=0\)

Step-by-Step Solution

Answer

The solution is \( y = \arctan(x) + 1 \).

Step 1: Recognize the Type of Differential Equation

This is a first-order ordinary differential equation with variables separable, of the form \( \frac{d y}{d x} = \frac{1}{1+x^{2}} \). The equation is not explicitly stated as a separable one, but it is directly integrable.

Step 2: Integrate both sides

To find the function \(y\), integrate both sides of \( \frac{d y}{d x} = \frac{1}{1+x^{2}} \). This leads to the integral \( y = \int \frac{1}{1+x^{2}} \, dx \). Recall that the integral of \( \frac{1}{1+x^{2}} \) is the arctangent function, \( \arctan(x) \), so \( y = \arctan(x) + C \), where \(C\) is the constant of integration.

Step 3: Apply Initial Conditions

Use the initial condition \(y_{0}=1\) when \(x_{0}=0\) to find \(C\). Substitute \(x=0\) and \(y=1\) into the equation: \[1 = \arctan(0) + C\]Since \(\arctan(0) = 0\), we have \(1 = C\). Thus, \(C = 1\).

Step 4: Write the Particular Solution

Incorporate the constant \(C\) back into the equation. The solution to the differential equation that satisfies the initial condition is \[y = \arctan(x) + 1\]

Key Concepts

First-Order Differential Equation

A first-order differential equation involves derivatives of a function, where the highest derivative is of the first degree. These are fundamental in describing many physical phenomena.

In our exercise, the given equation is \[ \frac{d y}{d x} = \frac{1}{1+x^{2}} \] which is characteristic of a first-order differential equation because it includes the first derivative \( \frac{d y}{d x} \) and equates it to a function of \(x\).

Understanding such equations is important because they help to describe things like velocity in physics or decay in radioactive substances. Solving these equations entails finding the original function \(y(x)\) from its derivative expression.

Initial Conditions

Initial conditions are values provided to specify the particular solution of a differential equation, eliminating the arbitrary constant. For differential equations like \[ \frac{d y}{d x} = \frac{1}{1+x^{2}} \] after solving it generally, we obtain a solution containing a constant \(C\).

To find \(C\), initial conditions, such as \(y_0=1\) when \(x_0=0\), are crucial. These conditions are like anchor points that ensure our solution fits the physical or geometrical context it models. Substituting these values into the integrated solution helps to determine the specific form of \(y(x)\).

Integration

Integration is a core mathematical operation used to solve differential equations by finding an antiderivative. Here's how it works with our equation:

• Start with the equation \( \frac{d y}{d x} = \frac{1}{1+x^{2}} \).
• Integration of the right-hand side gives \( y = \int \frac{1}{1+x^{2}} \, dx \).

This transformation moves from the derivative back to the original function. The integral of \( \frac{1}{1+x^{2}} \) is a known result, the arctangent or \( \arctan(x) \). Hence, the solution after integrating once is \[ y = \arctan(x) + C \], where \(C\) is determined using initial conditions.

Arctangent Function

The arctangent function, denoted \( \arctan(x) \), is an inverse trigonometric function, representing the angle whose tangent is \(x\). It's crucial in our context because the integral \( \int \frac{1}{1+x^{2}} \, dx \) directly simplifies to \( \arctan(x) \).

Understanding this function's role is important for solving our differential equation, as it allows us to obtain a precise expression for \(y(x)\). The arctangent is often involved in integrals related to circular trigonometric functions.

In our solution, substituting the initial condition into \[ y = \arctan(x) + C \] gives the particular solution \[ y = \arctan(x) + 1 \].