Problem 21 - Calculus: Early Transcendentals

Question

Solve the differential equation \( y' = x + y \) by making the change of variable \( u = x + y. \)

Step-by-Step Solution

Answer

The solution is \( y = -1 + Ce^x - x \).

Step 1: Substitute Variables

Given the differential equation, we need to make the substitution \( u = x + y \). Then, the derivative of \( y \) with respect to \( x \) is \( y' = u' - 1 \) because \( u = x + y \) implies \( u' = 1 + y' \). Substitute these into the original equation.

Step 2: Replace the Original Equation

By substituting the variables, the differential equation becomes \( u' - 1 = u \). Rearranging gives \( u' = u + 1 \). This is now a new differential equation in terms of \( u \).

Step 3: Solve the Linear Differential Equation

The equation \( u' = u + 1 \) can be rewritten as \( u' - u = 1 \). This is a first-order linear differential equation. The integrating factor is \( e^{-\int 1 \, dx} = e^{-x} \).

Step 4: Apply the Integrating Factor

Multiply the entire differential equation by the integrating factor: \( e^{-x}u' - e^{-x}u = e^{-x} \). This equation simplifies to \((e^{-x} u)' = e^{-x} \).

Step 5: Integrate Both Sides

Integrating both sides with respect to \( x \): \( \int (e^{-x} u)' \, dx = \int e^{-x} \, dx \). The left side integrates to \( e^{-x} u \) and the right side integrates to \( -e^{-x} + C \), where \( C \) is the constant of integration.

Step 6: Solve for \( u \)

Solving for \( u \): \( e^{-x} u = -e^{-x} + C \). Multiply through by \( e^x \) to get \( u = -1 + Ce^x \).

Step 7: Substitute Back to Original Variables

We have \( u = x + y = -1 + Ce^x \). Solving for \( y \), \( y = -1 + Ce^x - x \). This is the solution to the differential equation.

Key Concepts

Change of Variables

When solving complex differential equations, changing variables is a powerful technique that simplifies the problem. Suppose we have an equation that's challenging to solve in its current form. By replacing the variables with new ones, we can often make the equation much more manageable.
In our example, the differential equation given is \( y' = x + y \). To simplify, we use a substitution: let \( u = x + y \). This creates a new variable \( u \) that encapsulates the sum of \( x \) and \( y \).

• The original equation \( y' = x + y \) turns into a more tractable form after this substitution.
• It allows us to focus on solving a transformed equation without the linear complexity present in the original equation.
• Through such substitutions, many seemingly complex differential equations can be reduced to simpler forms.

Linear Differential Equations

A linear differential equation is one where the dependent variable and its derivatives appear linearly. This means no powers or products of the dependent variable and its derivatives occur. The equation \( u' - u = 1 \) is an example of a first-order linear differential equation.
In general, a first-order linear differential equation can be written as:

• \( y' + P(x)y = Q(x) \)
• In our case, \( P(x) = -1 \) and \( Q(x) = 1 \).
• These equations often require specific solution methods, such as using an integrating factor, because they cannot be solved through simple algebraic manipulation.

Integrating Factor

The integrating factor is a function that, when multiplied by a linear differential equation, transforms it into an exact equation. This makes it possible to solve the equation by direct integration.
For the equation \( u' - u = 1 \), the integrating factor is determined by:

• Calculating \( e^{\int -1 \, dx} = e^{-x} \).
• Multiplying the entire equation by \( e^{-x} \) simplifies the left-hand side into a derivative: \((e^{-x} u)'\).
• This transformation allows us to easily integrate and find a solution.

Integration

Integration is the reverse process of differentiation. Often, solving differential equations involves finding an antiderivative or integral to determine the solution. In the final steps of our solution, we integrated both sides of an equation:

• \( \int (e^{-x} u)' \, dx = \int e^{-x} \, dx \)
• The left-hand side integrates to \( e^{-x} u \), straightforward due to the integrating factor.
• The right-hand side yields \( -e^{-x} + C \), where \( C \) is the constant of integration.
• This integral operation is crucial for unlocking the solution \( u = -1 + Ce^x \).

Integration plays a key role in differential equations as it helps find functions whose derivatives result in given expressions or conditions.