Problem 22 - Calculus Early Transcendentals

Question

Find a solution to the initial-value problem. $$ y^{\prime \prime}+6 x=0, y(0)=1, y^{\prime}(0)=2 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Differential Equation

The given differential equation is \( y'' + 6x = 0 \). This is a second-order linear differential equation with constant coefficients and a non-homogeneous term \( 6x \).

Step 2: Solve the Homogeneous Equation

Consider the homogeneous version of the equation by setting the non-homogeneous term to zero: \( y'' = 0 \). Integrate once to get \( y' = C_1 \), and integrate again to get \( y = C_1x + C_2 \), where \( C_1 \) and \( C_2 \) are constants.

Step 3: Solve the Non-Homogeneous Equation

We use the method of undetermined coefficients to find a particular solution. Assume a particular solution of the form \( y_p = Ax^2 \). First, find \( y_p'' = 2A \). Substitute into the original equation: \( 2A + 6x = 0 \). Solving this gives \( A = -3x \), but since \( A \) must be a constant, first reevaluate the assumption to get the correct form by making \( A = -3/2 \).

Step 4: General Solution

Combine the homogeneous solution with the particular solution: \( y(x) = C_1 x + C_2 - \frac{3}{2} x^2 \).

Step 5: Apply Initial Conditions

Use the initial conditions to solve for \( C_1 \) and \( C_2 \). First, apply \( y(0) = 1 \):\[ C_2 = 1 \]Next, apply \( y'(0) = 2 \) by differentiating the general solution:\[ y'(x) = C_1 - 3x \]At \( x = 0 \), \( y'(0) = C_1 = 2 \).

Step 6: Final Solution

Now substitute \( C_1 = 2 \) and \( C_2 = 1 \) back into the general solution:\[ y(x) = 2x + 1 - \frac{3}{2} x^2 \]. This is the solution to the initial-value problem.

Key Concepts

Initial Value Problem

In mathematics, an initial value problem (IVP) is a type of differential equation that has additional conditions specified at the start of the process; these conditions are known as initial conditions. These problems involve finding a function that not only satisfies a differential equation but also meets the specific values at a given point, usually at the start of the function.

• The differential equation describes the rate at which something changes.
• The initial conditions provide specific starting details: for example, the value of the function and possibly its derivatives at a specific point.

In our original problem, the initial value conditions are given at \( y(0)=1 \) and \( y'(0)=2 \). These conditions help us find the specific solution out of all possible solutions of the differential equation that meets these criteria. Using the initial conditions means that we can solve for any constants introduced during the integration process. This is why the step towards obtaining the final solution in an initial value problem is applying these conditions to determine the constants.

Second-order Differential Equation

Second-order differential equations are equations involving a function and its derivatives up to the second order. A typical form of such an equation is \( y'' + p(x)y' + q(x)y = g(x) \), where \( y'' \) is the second derivative of \( y \), and \( p(x) \), \( q(x) \), and \( g(x) \) are functions of \( x \).

• The second-order derivative represents acceleration or the rate of change of the rate of change, crucial in physics and engineering.
• Solving these equations often involves transforming them into two first-order equations.

In the given exercise \( y'' + 6x = 0 \), we have a second-order differential equation without a first derivative term \( p(x) \) and where the non-homogeneous term \( g(x) \) is given as \( 6x \). Solving such equations involves finding the general solution to the associated homogeneous equation first, which only includes the homogeneous part (without \( g(x) \)), followed by solving for a particular solution that tackles the non-homogeneous part. The full solution becomes a sum of both the homogeneous and the particular solutions, incorporating any conditions provided, such as initial values.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions to non-homogeneous linear differential equations. This method works well when the non-homogeneous part, \( g(x) \), is a polynomial, exponential, sine, cosine, or a sum of these.

• The method involves guessing a form for the particular solution, \( y_p \), based on the type of non-homogeneous term \( g(x) \).
• We then determine the coefficients by plugging \( y_p \) into the differential equation and solving for these coefficients.

In the original exercise, we identified that the non-homogeneous term \( 6x \) suggests trying a polynomial form for the particular solution, \( y_p = Ax^2 \). After substituting into the equation and simplifying, we found that \( A = -\frac{3}{2} \), which completed our particular solution. Hence, the method of undetermined coefficients provides a structured pathway to find a specific solution to couple with the solution of the homogeneous equation to form the overall solution of the initial value problem.