Question

Find the general solution to the differential equations. $$ y^{\prime}=3 x-2 y $$

Short answer

The general solution is \( y(x) = \frac{3x}{2} - \frac{3}{4} + Ce^{-2x} \).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \( y' = 3x - 2y \). This is a first-order linear differential equation, which can be written in the standard form \( y' + P(x)y = Q(x) \) where \( P(x) = 2 \) and \( Q(x) = 3x \).

Find the Integrating Factor

To solve the linear differential equation, we need to find the integrating factor. The integrating factor, \( \mu(x) \), is given by \( e^{\int P(x) \, dx} \). Calculate \( \mu(x) = e^{\int 2 \, dx} = e^{2x} \).

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{2x} \)\[e^{2x}y' + 2e^{2x}y = 3xe^{2x} \]

Recognize the Left-Hand Side as a Derivative

The left-hand side of the equation \( e^{2x}y' + 2e^{2x}y \) can be expressed as the derivative of a product: \( \frac{d}{dx}(e^{2x}y) \). So the equation becomes: \[ \frac{d}{dx}(e^{2x}y) = 3xe^{2x} \]

Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^{2x}y) \, dx = \int 3xe^{2x} \, dx \]The left side integrates to \( e^{2x}y \), and the right side requires integration by parts.

Perform Integration by Parts

Using integration by parts on the right side, where \( u = 3x \) and \( dv = e^{2x} \, dx \) with \( du = 3 \, dx \) and \( v = \frac{1}{2}e^{2x} \), we find:\[ \int 3x e^{2x} \, dx = \frac{3x}{2}e^{2x} - \frac{3}{4}e^{2x} + C \]

Solve for \( y \)

Substitute back:\[ e^{2x}y = \frac{3x}{2}e^{2x} - \frac{3}{4}e^{2x} + C \]Solve for \( y \) by dividing through by \( e^{2x} \):\[ y = \frac{3x}{2} - \frac{3}{4} + Ce^{-2x} \]

Write the General Solution

Thus, the general solution to the differential equation is:\[ y(x) = \frac{3x}{2} - \frac{3}{4} + Ce^{-2x} \] where \( C \) is an arbitrary constant dependent on initial conditions.

Key concepts

Integrating Factor

To solve a first-order linear differential equation like the one given in the exercise, using an integrating factor is one of the key methods. The integrating factor helps transform the differential equation into one that is easily integrable. For any first-order linear differential equation of the form \( y' + P(x) y = Q(x) \), the integrating factor \( \mu(x) \) is calculated as \( e^{\int P(x) \, dx} \).

• First, identify the function \( P(x) \) from your equation. In our example, \( P(x) = 2 \).
• Calculate the integral: \( \int 2 \, dx = 2x \).
• Apply the exponential function to get your integrating factor: \( e^{2x} \).

With this integrating factor, multiply every term in the original differential equation to rewrite it in a format where the left side becomes the derivative of a product. This allows for straightforward integration, as the simplified side is automatically a derivative.

Integration by Parts

Once the left-hand side of the differential equation is a derivative, integration by parts is often required to solve the right-hand side. The integration by parts formula is based on the product rule for differentiation and is given by:\[\int u \, dv = uv - \int v \, du\]

• Choose \( u \) and \( dv \) from your integral carefully. For example, in \( \int 3x e^{2x} \, dx \), let \( u = 3x \) and \( dv = e^{2x} \, dx \).
• Differentiate \( u \) to find \( du = 3 \, dx \) and integrate \( dv \) to get \( v = \frac{1}{2}e^{2x} \).
• Substitute these into the integration by parts formula to evaluate the integral.

Thus, \( \int 3x e^{2x} \, dx \) becomes \( \frac{3x}{2}e^{2x} - \frac{3}{4}e^{2x} + C \), where \( C \) is the constant of integration.

General Solution of Differential Equations

The general solution of a first-order linear differential equation includes the expression derived from solving the equation, incorporating arbitrary constants that embody family solutions. After integrating and simplifying the differential equation using the integrating factor and possibly integration by parts, solve for \( y \).

• Using the example, after multiplying by \( e^{2x} \) and integrating, we reached \( e^{2x}y = \frac{3x}{2}e^{2x} - \frac{3}{4}e^{2x} + C \).
• To isolate \( y \), divide the entire equation by \( e^{2x} \), resulting in: \( y = \frac{3x}{2} - \frac{3}{4} + Ce^{-2x} \).
• \( C \) is an arbitrary constant. It gets defined by the initial conditions of a specific problem if provided.

This creates a family of solutions for the differential equation, represented by differing values of \( C \), allowing adaption to various initial conditions.