Problem 12 - Calculus for Biology and Medicine

Question

Find the general solution of the differential equations in Problems 1-12 using the method of integrating factors: $$ \frac{d y}{d x}+x=y $$

Step-by-Step Solution

Answer

The general solution is \( y(x) = -x - 1 + Ce^x \), where \( C \) is a constant.

Step 1: Identify the Equation Standard Form

Write the given differential equation \( \frac{dy}{dx} + x = y \) in the standard linear form: \( \frac{dy}{dx} + p(x) y = q(x) \). This can be rewritten as \( \frac{dy}{dx} - y = -x \) implying \( p(x) = -1 \) and \( q(x) = -x \).

Step 2: Find the Integrating Factor

Calculate the integrating factor \( \mu(x) \) using the formula \( \mu(x) = e^{\int p(x) \, dx} \). Here, integrating \( p(x) = -1 \) gives \( \mu(x) = e^{-x} \).

Step 3: Multiply Equation by Integrating Factor

Multiply every term of the differential equation by the integrating factor \( e^{-x} \): \( e^{-x} \frac{dy}{dx} - e^{-x}y = -xe^{-x} \).

Step 4: Recognize Left-hand Side as Derivative

The left side of the equation becomes a derivative of a product: \( \frac{d}{dx}(e^{-x}y) = -xe^{-x} \).

Step 5: Integrate Both Sides

Integrate the entire equation: \( \int \frac{d}{dx}(e^{-x}y) \, dx = \int -xe^{-x} \, dx \). The left-hand side simply becomes \( e^{-x}y \).

Step 6: Integrate the Right-hand Side

To integrate \( -xe^{-x} \), use integration by parts, letting \( u = x \), \( dv = -e^{-x}dx \). Then \( du = dx \), \( v = e^{-x} \). Applying integration by parts gives \( \int -xe^{-x} \, dx = -xe^{-x} + \int e^{-x} \, dx = -xe^{-x} - e^{-x} + C \).

Step 7: Solve for y(x)

The integrated equation is \( e^{-x}y = -xe^{-x} - e^{-x} + C \). Multiply every term by \( e^x \) to solve for \( y(x) \): \( y = -x - 1 + Ce^x \).

Step 8: State the General Solution

Thus, the general solution to the differential equation \( \frac{dy}{dx} + x = y \) is \( y(x) = -x - 1 + Ce^x \), where \( C \) is an arbitrary constant.

Key Concepts

Integrating Factors

When solving linear first-order differential equations, integrating factors play a crucial role. They are a technique that transforms non-exact equations into exact ones. But what exactly is an integrating factor? Simply put, it is a function, commonly denoted as \( \mu(x) \), that you multiply across the entire equation to facilitate its solution.

The integrating factor is calculated from the equation's linear form, expressed as \( \frac{dy}{dx} + p(x) y = q(x) \). The formula for the integrating factor is \( \mu(x) = e^{\int p(x) \, dx} \).

• Identifying \( p(x) \) from your equation is crucial.
• Compute the integral of \( p(x) \) with respect to \( x \).
• Exponentiate this result to obtain the integrating factor.

Once calculated, this factor helps transform the differential equation into a solvable form by recognizing a common derivative form on one side of the equation.

Linear Differential Equation

A linear differential equation takes a structured form that can be expressed as \( \frac{dy}{dx} + p(x) y = q(x) \). It's called "linear" because neither the function \( y \) nor its derivative \( \frac{dy}{dx} \) are raised to any power other than one. The linearity is specific concerning \( y \) and \( \frac{dy}{dx} \), not in terms of \( x \).

Understanding linear differential equations involves:

• Identifying the coefficient functions \( p(x) \) and \( q(x) \).
• Setting up your equation to match the standard form required for other solving techniques, like integrating factors.

Linear equations are easier to handle, primarily due to their structured nature, which allows for clear systematic approaches like the one involving integrating factors.

General Solution

A general solution to a differential equation represents a family of functions that satisfy the given equation. For a first-order linear differential equation, the solution typically includes arbitrary constants that represent this family.

Once you've applied the integrating factor method and integrated both sides, you arrive at a general solution form, which always includes a constant \( C \).

For example, the solution to \( \frac{dy}{dx} - y = -x \) becomes \( y(x) = -x - 1 + Ce^x \). Here, \( C \) is an arbitrary constant representing an infinite number of solutions. The actual solution depends on the initial conditions or additional constraints provided beyond the equation itself.

Integration by Parts

Integration by parts is an essential technique used to integrate products of two functions. When facing an integral that straightforward methods can't solve, integration by parts often helps. It's based on the product rule of differentiation. The formula is given by:\[ \int u \, dv = uv - \int v \, du \]
To employ this technique:

• Choose functions \( u \) and \( dv \) from the integral.
• Differentially differentiate \( u \) to get \( du \), and integrate \( dv \) to find \( v \).
• Apply the integration by parts formula to simplify the integral.

In our specific differential equation, integrating \( -xe^{-x} \) required us to choose \( u = x \) and \( dv = -e^{-x}dx \). This technique transformed the integral into a more manageable form, assisting in solving the differential equation.