Problem 50 - Calculus for Biology and Medicine

Question

Find the general solution of the differential equation. $$ \frac{d y}{d x}=e^{x+1}, x>0 $$

Step-by-Step Solution

Answer

The general solution is \( y = e^{x+1} + C \), where \( C \) is a constant.

Step 1: Identify the Type of Differential Equation

The equation \( \frac{dy}{dx} = e^{x+1} \) is a first-order ordinary differential equation. This equation can be solved through integration as it is a separable differential equation.

Step 2: Integrate with Respect to x

To find the function \( y(x) \), we integrate the right-hand side of the equation with respect to \( x \). We have:\[y = \int e^{x+1} \, dx.\]

Step 3: Simplify the Integral

The integral \( \int e^{x+1} \, dx \) can be simplified using substitution. Let \( u = x + 1 \), which means \( du = dx \). The integral becomes:\[y = \int e^u \, du.\]

Step 4: Solve the New Integral

The integral \( \int e^u \, du \) is straightforward. We know that \( \int e^u \, du = e^u + C \), where \( C \) is the constant of integration. Thus:\[y = e^u + C.\]

Step 5: Substitute Back to Original Variable

Replace \( u \) back with \( x + 1 \):\[y = e^{x+1} + C.\]This represents the general solution to the differential equation.

Key Concepts

Separable Differential Equations

Separable differential equations are a type of ordinary differential equation (ODE) where the variables can be separated on opposite sides of the equation. This means we can write the differential equation in a form where one variable, along with its differential, is on one side and the other variable with its differential is on the opposite side.

Let's address this with a simple example: consider the equation \( \frac{dy}{dx} = g(x) \cdot h(y) \). This can be rearranged to \( \frac{dy}{h(y)} = g(x) \cdot dx \). By integrating both sides, we find a potential solution for \( y \) in terms of \( x \) and possibly an arbitrary constant.

In our original exercise, the differential equation \( \frac{dy}{dx} = e^{x+1} \) was directly separable since it only involved \( x \). This means the integration of the right side will immediately lead to a solution for \( y \).

Key steps to approach such problems:

• Separate the variables.
• Integrate both sides.
• Substitute back any replacements made during the integration process.

Separable differential equations are fundamental and offer a straightforward pathway to solving first-order ODEs.

First-Order Ordinary Differential Equations

First-order ordinary differential equations are a class of ODEs where the highest derivative involved is the first derivative. These equations are crucial in modeling a range of real-world phenomenons, such as population growth, heat transfer, and motion.

The general form of a first-order ODE is \( \frac{dy}{dx} = f(x, y) \), where \( f \) might depend on both \( x \) and \( y \). Often, a specific analytical or numerical method is applied based on whether the differential equation is linear, separable, or of another special type.

In the provided exercise, the equation \( \frac{dy}{dx} = e^{x+1} \) is a first-order ODE. Here, \( f(x, y) = e^{x+1} \), showing that only \( x \) is involved, simplifying the problem greatly.

First-order ODEs can often be resolved when:

• The equation is linear or can be made linear through substitution.
• The equation is separable, allowing direct integration.

These equations form a basis for understanding more complex systems modeled by higher-order differential equations.

Integration Techniques

Integration techniques play a vital role in solving differential equations, especially when dealing with integration forms that arise from separating variables in equations. These techniques turn differential equations into simpler algebraic equations that can be more easily solved.

In our exercise, solving \( \frac{dy}{dx} = e^{x+1} \) required integrating \( e^{x+1} \). This was achieved by substitution, a core technique in integration. By letting \( u = x + 1 \), we changed the form of the integral to simple \( \int e^u \, du \), which solved directly as \( e^u + C \).

Common integration techniques include:

• Substitution: Useful when the integral can be transformed into a simpler form. Involves a change of variables to help directly integrate a function.
• Integration by parts: Helps when dealing with product integrals.
• Partial fraction decomposition: Useful for rational functions, breaking them into simpler parts that are easier to integrate.

Mastering these techniques is essential for anyone tackling differential equations, as they help unlock the solutions to seemingly complex integral problems.