Question

Find solutions \(y_{1}\) and \(y_{2}\) of the equation \(y^{\prime \prime}=0\) that satisfy the initial conditions $$y_{1}\left(x_{0}\right)=1, \quad y_{1}^{\prime}\left(x_{0}\right)=0 \quad \text { and } \quad y_{2}\left(x_{0}\right)=0, \quad y_{2}^{\prime}\left(x_{0}\right)=1 .$$ Then use Exercise 37 (c) to write the solution of the initial value problem $$y^{\prime \prime}=0, \quad y(0)=k_{0}, \quad y^{\prime}(0)=k_{1}$$ as a linear combination of \(y_{1}\) and \(y_{2}\).

Short answer

To summarize, in order to solve the given initial value problem, we determined the general solution of the equation \(y'' = 0\) to be \(y(x) = C_1x + C_2\). After applying the initial conditions for \(y_1\) and \(y_2\), we found the solutions \(y_1(x) = 1\) and \(y_2(x) = x - x_0\). The solution of the initial value problem as a linear combination of \(y_1\) and \(y_2\) is given by \(y(x) = a_1 + k_1(x - x_0)\), where \(a_1\) and \(k_1\) are constants found using the given initial conditions.

Step-by-step solution

Find the general solution of the equation y''=0

To find the general solution, we will integrate the equation \(y^{\prime \prime}=0\) twice: First integration (Integrate with respect to x): $$ y^{\prime} = \int y^{\prime \prime} \mathrm{d}x = \int 0 \mathrm{d}x = C_1 $$ Second integration (Integrate with respect to x): $$ y = \int y^{\prime} \mathrm{d}x = \int C_1 \mathrm{d}x = C_1x + C_2 $$ So, the general solution of the equation \(y^{\prime \prime}=0\) is: $$ y(x) = C_1x + C_2 $$

Apply the initial conditions for y1 and y2

Now, we will apply the given initial conditions for \(y_1\) and \(y_2\) to find the constants \(C_1\) and \(C_2\): Initial conditions for y1: $$ y_1(x_0) = 1 \implies C_1x_0 + C_2 = 1 $$ $$ y_1^{\prime}(x_0) = 0 \implies C_1 = 0 $$ Initial conditions for y2: $$ y_2(x_0) = 0 \implies C_1x_0 + C_2 = 0 $$ $$ y_2^{\prime}(x_0) = 1 \implies C_1 = 1 $$

Determine the solutions y1 and y2

Using the initial conditions, we can determine the solutions \(y_1\) and \(y_2\): For \(y_1\): $$ C_1 = 0 $$ $$ y_1(x) = C_2 $$ Since \(y_1(x_0) = 1\), it implies \(C_2 = 1\), and the solution is: $$ y_1(x) = 1 $$ For \(y_2\): $$ C_1 = 1 $$ $$ y_2(x) = x + C_2 $$ Since \(y_2(x_0) = 0\), it implies \(C_2 = -x_0\) and the solution is: $$ y_2(x) = x - x_0 $$

Write the solution of the initial value problem as a linear combination of y1 and y2

Now, we will use Exercise 37 (c) to write the solution of the initial value problem as a linear combination of \(y_1\) and \(y_2\). The general solution is given by: $$ y(x) = a_1y_1(x) + a_2y_2(x) $$ where \(a_1\) and \(a_2\) are constants. From the initial conditions given: $$ y(0) = k_0 \implies a_1y_1(0) + a_2y_2(0) = k_0 $$ $$ y^{\prime}(0) = k_1 \implies a_1y_1^{\prime}(0) + a_2y_2^{\prime}(0) = k_1 $$ We know from previous steps that \(y_1(x) = 1\), \(y_2(x) = x - x_0\), and their derivatives are \(y_1^{\prime}(x) = 0\), \(y_2^{\prime}(x) = 1\), respectively. Therefore, the equations become: $$ a_1 + a_2(-x_0) = k_0 $$ $$ a_2 = k_1 $$ Now we can express the general solution with respect to \(k_0\), \(k_1\), and \(x_0\): $$ y(x) = a_1(1) + a_2(x - x_0) = a_1 + k_1(x - x_0) $$ This is the solution of the initial value problem as a linear combination of \(y_1\) and \(y_2\).

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific conditions, called initial conditions, which must be satisfied by the solution. These conditions are given at the start of the problem, specifying the value of the solution and usually its derivatives at a particular point. In problems involving differential equations, initial value conditions play a critical role because they determine the specific solution among a family of possibilities.

In the given exercise, the equation is a second-order differential equation: \(y'' = 0\). This means the second derivative of the solution, \(y\), is zero everywhere, which significantly simplifies the problem. Here are the initial conditions applied:

• For \(y_1\), \(y_1(x_0) = 1\) and \(y_1'(x_0) = 0\)
• For \(y_2\), \(y_2(x_0) = 0\) and \(y_2'(x_0) = 1\)

These initial conditions help us find particular solutions, \(y_1\) and \(y_2\), that satisfy the differential equation at \(x_0\). Without specifying these conditions, we would only have a general solution that includes arbitrary constants.

General Solution

The general solution of a differential equation represents a family of functions that satisfy the equation. For second-order differential equations like \(y'' = 0\), the general solution involves two arbitrary constants because two integrations are required.

Upon integrating \(y'' = 0\):

• The first integration gives \(y' = C_1\) where \(C_1\) is an arbitrary constant.
• The second integration results in \(y = C_1x + C_2\), with another constant \(C_2\).

Therefore, the general solution here is \(y(x) = C_1x + C_2\). This form shows that the solution is a straight line whose slope is determined by \(C_1\), and the y-intercept is determined by \(C_2\). The initial conditions will dictate particular values for these constants, thus identifying a specific member of this family of solutions.

Linear Combination

A linear combination involves forming a new function by multiplying given functions by constants and then adding the results together. It is a powerful technique in solving differential equations, especially when dealing with superposition. Superposition is applicable in linear scenarios where the sum of particular solutions also satisfies the differential equation.

In the exercise, a solution of the initial value problem is written as a linear combination of the functions \(y_1\) and \(y_2\). This means combining these solutions as:

• \(y(x) = a_1y_1(x) + a_2y_2(x)\)

Here, \(a_1\) and \(a_2\) are constants determined by the initial conditions \(y(0) = k_0\) and \(y'(0) = k_1\).

By substituting in the values of \(y_1\) and \(y_2\) and their derivatives, we can solve for \(a_1\) and \(a_2\) to find the particular solution to the problem.

Second Order Differential Equation

A second-order differential equation is one that involves the second derivative of an unknown function. It might also contain lower-order derivatives and the function itself. These equations are prevalent in physics and engineering because they can describe a wide range of phenomena, from mechanical vibrations to electrical circuits.

The general form of a second-order differential equation is \(a(x)y'' + b(x)y' + c(x)y = f(x)\). In our exercise, it's simplified to \(y'' = 0\). This states that the second derivative is zero, implying the function must be a linear expression \(y = C_1x + C_2\) after integration.

The simplicity of \(y'' = 0\) allows us to easily calculate solutions using basic integration and apply initial conditions to find specific values of \(C_1\) and \(C_2\). It also demonstrates fundamental concepts like how initial conditions influence specific solutions in the context of a general solution family.