Problem 20 - Thomas Calculus in SI Units

Question

Solve the differential equations. $$\frac{d y}{d x}=x y+3 x-2 y-6$$

Step-by-Step Solution

Answer

The solution is \( y = \left(\frac{3}{4} + C e^{-2x}\right)x - \frac{9}{4} \).

Step 1: Rearrange the Equation

Start by rearranging the differential equation, \( \frac{d y}{d x} = x y + 3x - 2y - 6 \), so all terms involving \(y\) are on one side and the rest are on the other side. This becomes: \( \frac{d y}{d x} + 2y = xy + 3x - 6 \).

Step 2: Find the Integrating Factor

This is a linear first-order differential equation of the form \( \frac{d y}{d x} + P(x)y = Q(x) \). Here, \( P(x) = 2 \). The integrating factor is found by \( e^{\int P(x) \, dx} = e^{\int 2 \, dx} = e^{2x} \).

Step 3: Multiply by the Integrating Factor

Multiply through the entire equation by the integrating factor \( e^{2x} \): \( e^{2x} \frac{d y}{d x} + 2e^{2x}y = e^{2x}(xy + 3x - 6) \). This simplifies to \( \frac{d}{dx}(e^{2x}y) = e^{2x}xy + 3xe^{2x} - 6e^{2x} \).

Step 4: Integrate Both Sides

Integrate both sides with respect to \(x\):\( \int \frac{d}{dx}(e^{2x}y) \, dx = \int (e^{2x}xy + 3xe^{2x} - 6e^{2x}) \, dx \).The left side simplifies to \( e^{2x}y \).Next, divide the right side into parts and integrate each separately using integration by parts for \( e^{2x}xy \) and basic integration rules for the rest.

Step 5: Solve the Integrals

For the integral \( \int e^{2x}xy \, dx \), use integration by parts: let \( u = x \) and \( dv = e^{2x} \), then \( du = dx \) and \( v = \frac{1}{2}e^{2x} \). This gives \( \frac{x}{2}e^{2x} - \int \frac{1}{2}e^{2x} \, dx = \frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C \). For \( \int 3xe^{2x} \, dx \) and \( \int -6e^{2x} \, dx \), apply integration by parts and direct integration respectively.

Step 6: Combine the Results

Summing the parts, we find:\( e^{2x}y = \left(\frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C\right) + (\text{solution of } 3xe^{2x}) - 3e^{2x} + C_1 \). The first integral has an additional \( \frac{3}{4}e^{2x} \) from the second integral using parts:\( e^{2x}y = \left(\frac{3}{4}e^{2x} + 3\right)x - \frac{9}{4}e^{2x} + C \).

Key Concepts

Integrating Factor

An integrating factor is a useful tool used to solve linear first-order differential equations. It transforms a non-exact differential equation into an exact one, making it easier to solve. To use it, we first need to recognize the differential equation is in the form

• \( \frac{dy}{dx} + P(x)y = Q(x) \)
• Where \( P(x) \) and \( Q(x) \) are functions of \( x \).

The integrating factor, \( \mu(x) \), is given as \( e^{\int P(x) \, dx} \).
This transforms the original equation so that the left side can be written as a derivative of a product: \( \frac{d}{dx}(\mu(x)y) = \mu(x)Q(x) \).
In the given problem, the integrating factor is \( e^{2x} \). Using this factor, we multiply each term of the differential equation, allowing us to rewrite the equation with a simpler integration step.

Linear First-Order Differential Equation

Understanding linear first-order differential equations is key to applying methods like integrating factors. These equations typically look like:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)
• Where \( P(x) \) and \( Q(x) \) depend on \( x \) alone.

Linear equations are defined by their simple arithmetic operations on the main variable and its derivative.
To solve such equations, rearranging terms is the first step, followed by applying the integrating factor to simplify the solving process.
In this exercise, we began by shifting all "\( y \)" terms to one side leading to a more distinct representation of the problem.
This then facilitated the use of an integrating factor, setting the stage for solving through straightforward integration.

Integration by Parts

Integration by parts is a fundamental technique of calculus used to solve integrals, particularly when the integrand is a product of two functions. Its formula is derived from the product rule for differentiation and is given by:

• \[ \int u \, dv = uv - \int v \, du \]
• Where "\( u \)" is chosen from one part of the equation, while "\( dv \)" represents the remainder.

In our problem, integration by parts was used to tackle the integral \( \int e^{2x}xy \, dx \). By designating \( u = x \) and \( dv = e^{2x} \, dx \), we facilitate the integration process via differentiation of \( u \) and integration of \( dv \).
After applying integration by parts, the expression simplifies, allowing easier subsequent integration.
This method is repeatedly invaluable for integrals that involve products of polynomial, exponential, and trigonometric functions, helping transform complex integrals into manageable calculations.