Question

Solve the initial-value problem $$ y^{\prime}+3 y=2 x-1, \quad y(0)=3 $$

Short answer

The solution is \( y = \frac{2x - 1}{3} - \frac{2}{9} + \frac{32}{9}e^{-3x} \).

Step-by-step solution

Recognize the type of equation

The given differential equation is a first-order linear ordinary differential equation of the form \( y' + Py = Q \), where \( P = 3 \) and \( Q = 2x - 1 \).

Find the Integrating Factor

To solve the equation, we first find the integrating factor, \( \mu(x) \), given by \( e^{\int P \, dx} = e^{\int 3 \, dx} = e^{3x} \).

Multiply the entire equation by Integrating Factor

Multiply the whole differential equation \( y' + 3y = 2x - 1 \) by the integrating factor \( e^{3x} \), resulting in:\[ e^{3x}y' + 3e^{3x}y = (2x - 1)e^{3x} \]

Recognize as a derivative of product

The left-hand side of the equation: \( \frac{d}{dx}(e^{3x}y) = e^{3x}y' + 3e^{3x}y \), can be rewritten using the integrating factor, yielding:\[ \frac{d}{dx}(e^{3x}y) = (2x - 1)e^{3x} \]

Integrate both sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^{3x}y) \, dx = \int (2x - 1)e^{3x} \, dx \] This results in:\[ e^{3x}y = \int (2x - 1)e^{3x} \, dx + C \]

Solve the right-hand side

Using integration by parts, let \( u = 2x - 1 \) and \( dv = e^{3x} \, dx \). Then \( du = 2 \, dx \) and \( v = \frac{1}{3}e^{3x} \). Thus the integral becomes:\[ \int (2x - 1)e^{3x} \, dx = (2x - 1)\frac{1}{3}e^{3x} - \int \frac{2}{3}e^{3x} \, dx \] Calculate:\[ \int \frac{2}{3}e^{3x} \, dx = \frac{2}{9}e^{3x} \] Therefore:\[ \int (2x - 1)e^{3x} \, dx = \frac{1}{3}e^{3x}(2x - 1) - \frac{2}{9}e^{3x} \]

Write the general solution

Using the result of the integration from the previous step, the solution becomes:\[ e^{3x}y = \frac{1}{3}e^{3x}(2x - 1) - \frac{2}{9}e^{3x} + C \] Simplify the equation:\[ y = \frac{2x - 1}{3} - \frac{2}{9} + Ce^{-3x} \]

Apply initial condition

Use the initial condition \( y(0) = 3 \) to find \( C \):\[ 3 = \frac{2 \times 0 - 1}{3} - \frac{2}{9} + C \] \[ 3 = -\frac{1}{3} - \frac{2}{9} + C \] Calculate and solve for \( C \):\[ 3 = -\frac{3}{9} - \frac{2}{9} + C = -\frac{5}{9} + C \] \[ C = 3 + \frac{5}{9} = \frac{32}{9} \]

Write the final solution

Substitute \( C \) back into the expression for \( y \):\[ y = \frac{2x - 1}{3} - \frac{2}{9} + \frac{32}{9}e^{-3x} \] Simplify if necessary.

Key concepts

Integrating Factor

An integrating factor is a strategic tool used to solve first-order linear differential equations. It simplifies the equation, making it easier to solve. When confronting an equation of the form \( y' + Py = Q \), the integrating factor is often denoted as \( \mu(x) \) and takes the form \( e^{\int P \, dx} \). This factor converts the left-hand side of the differential equation into the derivative of a product, which can then be easily integrated.

In our exercise, the given differential equation is \( y' + 3y = 2x - 1 \). We identify \( P = 3 \) from this equation. Therefore, the integrating factor becomes \( e^{\int 3 \, dx} = e^{3x} \). By multiplying the entire equation by this integrating factor, we transform the differential equation into a form where the left-hand side is a full derivative: \( \frac{d}{dx}(e^{3x}y) \). This step is crucial as it paves the way for further integration and eventual solution of the equation.

Initial-Value Problem

An initial-value problem (IVP) combines a differential equation with a specified value, or condition, at a starting point. This condition, often given as \( y(x_0) = y_0 \), allows us to find the particular solution to the differential equation that passes through this specific point.

For our problem, the differential equation \( y' + 3y = 2x - 1 \) comes with the initial condition \( y(0) = 3 \). After solving the differential equation, we substitute \( x = 0 \) into the general solution to find the constant \( C \). The result is a specific solution that not only solves the differential equation but also satisfies the initial condition. For example, using \( y(0) = 3 \) in our case, we substituted into our derived solution to solve for \( C \) and find that \( C = \frac{32}{9} \). This results in a solution that perfectly fits the initial condition and meets the requirement of the given problem.

Integration by Parts

Integration by parts is a powerful technique used when integrating the product of two functions. It is the integral version of the product rule for differentiation and is especially useful when such products are involved. The formula is given by:\[\int u \, dv = uv - \int v \, du\]This method helps when direct integration of a product is difficult.

In our step-by-step solution, after finding the integrating factor and rewriting the differential equation, we reached a point where we needed to integrate \( \int (2x - 1)e^{3x} \, dx \). To tackle this, we let \( u = 2x - 1 \) and \( dv = e^{3x} \, dx \). After computing \( du = 2 \, dx \) and \( v = \frac{1}{3}e^{3x} \), we applied the integration by parts formula:

• First, compute \( uv \), which gives us \( (2x - 1) \frac{1}{3} e^{3x} \).
• Then, calculate the integral \( \int (\frac{2}{3}e^{3x}) \, dx \).
• Finally, substitute these results back to get the full solution.

Using this systematic method, we're able to evaluate the integral accurately and find the correct solution to the differential equation.