Problem 49 - Calculus for Biology and Medicine

Question

In Problems, solve each differential equation with the given initial condition. $$ \frac{d y}{d x}=(y+1) e^{-x}, \text { with } y(0)=2 $$

Step-by-Step Solution

Answer

The solution to the differential equation is \( y = 3e^{1 - e^{-x}} - 1 \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{dy}{dx} = (y + 1) e^{-x} \). This is a first-order linear differential equation. Since it can be separated, we can use the method of separation of variables to solve it.

Step 2: Separate Variables

Rewrite the equation to separate the variables: \( \frac{dy}{(y+1)} = e^{-x} \, dx \). This allows us to integrate both sides independently with respect to their variables.

Step 3: Integrate Both Sides

Integrate both sides of the equation: \( \int\frac{dy}{(y+1)} = \int e^{-x} \, dx \). This gives us: \( \ln|y+1| = -e^{-x} + C \), where \( C \) is the constant of integration.

Step 4: Solve for the Constant Using Initial Conditions

Use the initial condition \( y(0) = 2 \) to find \( C \). Substitute \( x = 0 \) and \( y = 2 \) into the equation: \( \ln|2+1| = -e^{0} + C \) which simplifies to \( \ln 3 = -1 + C \). Solving for \( C \), we get \( C = \ln 3 + 1 \).

Step 5: Solve for \( y \)

Substitute \( C \) back into the equation to find \( y \): \( \ln|y+1| = -e^{-x} + \ln 3 + 1 \). Exponentiate both sides to remove the natural log: \( |y+1| = e^{\ln 3 + 1 - e^{-x}} \). This simplifies to \( y+1 = 3e^{1 - e^{-x}} \). Thus, \( y = 3e^{1 - e^{-x}} - 1 \).

Key Concepts

Understanding Initial Conditions

In the realm of differential equations, initial conditions play a crucial role in determining the specific solution to a problem. They specify the value of the function (often denoted as \( y \)) at a particular point, typically at \( x = 0 \), which aids in finding the unique solution from a family of solutions.
In our example, the initial condition provided is \( y(0) = 2 \). This tells us that when \( x = 0 \), the value of \( y \) must be 2.
This condition is essential for calculating the constant of integration \( C \) when we solve the differential equation, thereby tailoring the solution specifically to the given scenario.

• Without these initial conditions, the solution could have infinitely many possibilities, each differing by a constant.
• Initial conditions essentially "anchor" the equation, helping to pinpoint a precise path for the solution curve.

What is a First-Order Linear Differential Equation?

A first-order linear differential equation is among the most straightforward types of differential equations you'll encounter. It involves the first derivative of the function \( y \) and can be written in the standard form: \( \frac{dy}{dx} + P(x)y = Q(x) \).
In the problem we tackled, the equation \( \frac{dy}{dx} = (y + 1) e^{-x} \) can technically be classified under this category since it primarily deals with the first derivative of \( y \).

• These equations are called "linear" because the term involving \( y \) is to the power of one (hence linearity).
• Being first-order implies that it involves only the first derivative, \( \frac{dy}{dx} \).
• Such equations are often solvable using straightforward methods like integrating factors or separation of variables.

The Method of Separation of Variables

Separation of variables is a powerful technique for solving first-order differential equations that can be rearranged into a product of functions, each dependent on a different variable. Our given equation is \( \frac{dy}{dx} = (y + 1)e^{-x} \). By using separation of variables, we rewrite it as \( \frac{dy}{(y+1)} = e^{-x} \, dx \).
This transformation allows each side of the equation to be integrated independently: one side with respect to \( y \) and the other with respect to \( x \).

• Integrating both sides individually leads to a general solution involving an integration constant \( C \).
• This method is especially efficient for equations that permit easy isolation of each variable on either side.
• After finding the general solution, initial conditions are used to determine the particular value of \( C \) for a specific solution.

Separation of variables is thus integral to yielding a solution that fits our initial condition, making it highly practical for certain forms of differential equations.