Question

For the following problems, find the general solution to the differential equation.\(y^{\prime}=4^{x}\)

Short answer

The general solution is \( y(x) = \frac{4^x}{\ln 4} + C \).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is a first-order linear differential equation: \( y' = 4^x \). Our goal is to find a function \( y(x) \) such that its derivative \( y'(x) \) is equal to \( 4^x \).

Set Up the Integral Equation

To find \( y(x) \), we need to integrate both sides of the equation. We write this as: \( \int y' \, dx = \int 4^x \, dx \).

Solve the Integral

The right side of the equation requires solving the integral of \( 4^x \). Recall the integral formula for an exponential function: \( \int a^x \, dx = \frac{a^x}{\ln a} + C \), where \( a \) is a constant. For \( a = 4 \), we have: \( \int 4^x \, dx = \frac{4^x}{\ln 4} + C \).

Write the General Solution

After integrating, we obtain the general solution: \( y(x) = \frac{4^x}{\ln 4} + C \), where \( C \) is the constant of integration.

Key concepts

First-Order Differential Equation

A first-order differential equation involves derivatives of the first order but no higher derivatives. This means it contains terms like \( y' \) or \( dy/dx \). In our exercise, the equation \( y' = 4^x \) is a first-order differential equation because it involves the first derivative of \( y \) with respect to \( x \).

The primary aim when solving these equations is to find a function \( y(x) \) that satisfies the relationship for all values of \( x \). First-order differential equations can often be solved through the process of integration, as it involves finding the antiderivative of a function. This method is particularly useful for equations where the derivative \( y' \) is expressed as a function of \( x \) alone.

In this particular problem, because \( y' = 4^x \) directly gives us an expression for the derivative, we can proceed by integrating both sides to find the general solution. This is a common approach when dealing with first-order differential equations of this form.

Integration

Integration is a fundamental concept in calculus used to find antiderivatives—a process that essentially reverses differentiation. It allows us to determine the original function given its derivative.

In the context of solving the differential equation \( y' = 4^x \), integration is used to find \( y(x) \) from the given derivative. We set up the problem as:\[\int y' \, dx = \int 4^x \, dx\]
Since integration undoes the process of differentiation, the left side of the equation simplifies to \( y \), while the right side must be integrated with respect to \( x \).

This process involves finding a function whose derivative is \( 4^x \). Often, integrations involve constants, and after integration, an adjusting constant, usually noted as \( C \), is added because indefinite integrals reflect a family of functions that differ by a constant. In our solution, we found that integrating \( 4^x \) resulted in \( y(x) = \frac{4^x}{\ln 4} + C \). The constant \( C \) represents any vertical shift of the function and acknowledges that derivatives of constant terms are zero.

Exponential Functions

Exponential functions are a key part of mathematics and calculus, particularly when dealing with growth and decay rates. These functions have the form \( a^x \), where \( a \) is a constant and \( x \) is the exponent.

In many calculus problems, exponential functions are involved, especially when dealing with integrals and differential equations. When solving \( y' = 4^x \), the key step was integrating an exponential function.

The integration formula for exponential functions is:\[\int a^x \, dx = \frac{a^x}{\ln a} + C\]This formula indicates that the integral of an exponential function can be expressed in terms of itself, divided by the natural logarithm of the base \( a \).

In our exercise, \( a = 4 \), thus resulting in \( \int 4^x \, dx = \frac{4^x}{\ln 4} + C \). Recognizing how to work with exponential terms simplifies many calculus operations, especially when dealing with their growth behavior as they rapidly increase or decrease based on the base value.