Problem 18 - Calculus Single Variable

Question

Solve the given differential equation. $$ \frac{d y}{d x}=y \sec (x) $$

Step-by-Step Solution

Answer

The solution to the differential equation is \( y = C(\sec(x) + \tan(x)) \).

Step 1: Separation of Variables

The given differential equation is \( \frac{d y}{d x} = y \sec(x) \). Start by using the separation of variables technique. Divide both sides by \( y \) to isolate \( dy \) on one side: \( \frac{1}{y} dy = \sec(x) dx \).

Step 2: Integrate Both Sides

Integrate both sides of the equation. The left side becomes \( \int \frac{1}{y} dy \) which is \( \ln|y| + C_1 \). The right side is \( \int \sec(x) dx \), which is \( \ln|\sec(x) + \tan(x)| + C_2 \).

Step 3: Exponentiate to Solve for y

To solve for \( y \), exponentiate both sides to remove the natural logarithm: \( |y| = e^{C_1} e^{\ln|\sec(x) + \tan(x)| + C_2} = Ce^{\ln|\sec(x) + \tan(x)|} \). Simplify to \( y = C(\sec(x) + \tan(x)) \), where \( C = e^{C_1 + C_2} \).

Key Concepts

Separation of Variables

One of the most effective methods for solving differential equations is the separation of variables. This is useful when a differential equation can be expressed as a product of functions, each depending solely on one variable. In our exercise, the equation given is \( \frac{d y}{d x} = y \sec(x) \). By using the separation of variables technique, you separate all terms in one variable (\( y \)) on one side of the equation and all terms in the other variable (\( x \)) on the other.

• Start with separating the variables: divide both sides by \( y \).
• You'll get \( \frac{1}{y} dy = \sec(x) dx \).

This prepares the differential equation for integration by isolating the variables.

Integration

Integration is the process used to find the function that describes the rate of change, effectively helping in finding the antiderivative or original function. In the separated equation \( \frac{1}{y} dy = \sec(x) dx \), integration is the next logical step.

• The left side, \( \int \frac{1}{y} dy \), integrates to \( \ln|y| + C_1 \). Note: \( C_1 \) is an integration constant.
• The right side, \( \int \sec(x) dx \), integrates to \( \ln|\sec(x) + \tan(x)| + C_2 \), where \( C_2 \) is another constant.

By integrating, we find the expressions that were originally differentiated, allowing us to solve the equation.

Exponentiation

After integrating, we often have natural logarithms like \( \ln|y| \) and \( \ln|\sec(x) + \tan(x)| \). To simplify these expressions and solve for \( y \), exponentiation is key.

• Exponentiating both sides removes the logarithm, turning the expression into \( |y| = e^{\text{right side}} \).
• Thus, \( |y| = e^{C_1} \cdot e^{\ln|\sec(x) + \tan(x)| + C_2} \).
• This simplifies to \( |y| = C |\sec(x) + \tan(x)| \), where \( C = e^{C_1 + C_2} \).

Exponentiation here helps transition from a logarithmic expression back to an algebraic form, allowing for further simplification.

Calculus

Differential equations are an integral part of calculus, involving both differentiation and integration. Calculus handles continuous change, and differential equations model this life of function with respect to other variables.

• These equations incorporate functions and their derivatives, showcasing how variables are interdependent.
• The primary techniques in calculus—differentiation and integration—are used in tandem to solve problems like the one presented.

This specific problem demonstrates the interconnectedness of calculus concepts, emphasizing the role of calculus in interpreting and solving real-world situations by understanding the dynamics of changing quantities.