Problem 21 - Calculus Early Transcendentals

Question

Find a solution to the initial-value problem. $$ y^{\prime}+4 x=2, y(0)=3 $$

Step-by-Step Solution

Answer

The solution is \( y = x + 3 \).

Step 1: Identify the Problem Type

The given problem is a first-order linear ordinary differential equation with an initial condition.

Step 2: Rearrange the Equation

Start with the differential equation: \( y' + 4x = 2 \). Rearrange it into the standard form for linear equations: \( y' = -4x + 2 \).

Step 3: Solve the Homogeneous Equation

The homogeneous part of the equation is \( y' = 0 \), which yields a constant solution, say \( y_h = C \).

Step 4: Find a Particular Solution

A particular solution can be found by guessing a constant solution since \( y' = -4x + 2 \). Substitute \(y = k\) into the differential equation: \( 0 + 4x = 2 \) gives \( k' = 2 - 4x \). Thus, \( y_p = x \) is a particular solution.

Step 5: General Solution

The general solution is a combination of the homogeneous and particular solutions: \( y = y_h + y_p = C + x \).

Step 6: Apply Initial Condition

Use the initial condition given: \( y(0) = 3 \). Substitute into the general solution: \( 3 = C + 0 \). Solve for \( C \) to get \( C = 3 \).

Step 7: Write the Final Solution

Substitute \( C = 3 \) back into the general solution to get the specific solution that satisfies the initial condition: \( y = 3 + x \).

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is an equation of the form \( y' + p(x)y = q(x) \), where \( y' \) is the derivative of \( y \) with respect to \( x \), and \( p(x) \) and \( q(x) \) are functions of \( x \). These equations are called "first-order" because \( y' \) is the highest derivative in the equation.

For these types of equations, the integrating factor method is commonly used to find solutions. However, in simple cases where \( p(x) \) is zero, like in our example \( y' + 4x = 2 \), one can directly rearrange and simplify the equation to make it easier to solve.

This initial-value problem also specifies \( y(0) = 3 \), meaning the value of the solution function \( y(x) \) is known at \( x = 0 \).

Homogeneous Solution

The homogeneous solution involves solving the differential equation's homogeneous part, which is derived by setting the non-homogeneous term to zero. In simpler terms, for the equation \( y' + 4x = 2 \), you first consider the equation \( y' + 4x = 0 \).

Since the original equation depends on \( x \), not \( y \), you can think about the homogeneous solution by solving \( y' = 0 \). The solution to this is a constant, \( C \), because the derivative of a constant is zero.

This constant solution \( C \) forms the cornerstone of the general solution because it represents the arbitrary constants that appear when solving differential equations.

Particular Solution

Finding a particular solution means finding one specific solution to the non-homogeneous ordinary differential equation. This is a solution of the form \( y = y_p \) that satisfies only the non-homogeneous part of the equation.

In the case of \( y' + 4x = 2 \), because there is no \( y \)-term on the left, a simple guess could be that \( y_p = x \), meeting the right-hand side's linearity, \( 2 \). When substituting back into the equation, it satisfies the differential equation, making it a viable particular solution.

The particular solution is not influenced by the initial condition and provides just one curve out of the family of curves given by the general solution.

General Solution

The general solution of a first-order linear differential equation is formed by combining the homogeneous solution and the particular solution. It represents all possible solutions of the differential equation.

Once you have the homogeneous solution \( y_h = C \) and the particular solution \( y_p = x \), you can combine them to form the general solution: \( y = y_h + y_p = C + x \). This solution incorporates the arbitrary constant \( C \) from the homogeneous solution that adapts based on any initial conditions.

Finally, we use the initial condition, \( y(0) = 3 \), to determine the specific value of \( C \). Substituting \( x = 0 \) into the general solution gives us \( 3 = C + 0 \), thus \( C = 3 \). The corresponding specific solution is \( y = 3 + x \), which satisfies both the differential equation and the initial condition.