Question

In each of Problems 1 through 4 determine \(\phi^{\prime \prime}\left(x_{0}\right), \phi^{\prime \prime \prime}\left(x_{0}\right),\) and \(\phi^{\mathrm{iv}}\left(x_{0}\right)\) for the given point \(x_{0}\) if \(y=\phi(x)\) is a solution of the given initial value problem. $$ y^{\prime \prime}+x y^{\prime}+y=0 ; \quad y(0)=1, \quad y^{\prime}(0)=0 $$

Short answer

Answer: The second, third, and fourth derivatives of the solution φ(x) evaluated at x₀=0 are: φ''(0) = -1, φ'''(0) = 0, and φ^(4)(0) = 1.

Step-by-step solution

Find the general solution

The differential equation is given as: $$ y''+xy'+y=0 $$ To find the general solution, we can use the method of variation of parameters. First, find the complementary function \(y_c(x)\), which is the solution of the homogeneous equation: $$ y''+xy'+y=0 $$ Since this is a linear homogeneous differential equation, we can try a solution of the form \(y=e^{rx}\). Plugging this into the equation, we obtain the characteristic equation: $$ r^2 + r + 1=0 $$ This is a quadratic equation with no real roots, which means that the general solution of the homogeneous equation is an exponential function of the form: $$ y_c(x)=Ae^{r_1x} + Be^{r_2x} $$ where \(r_1\) and \(r_2\) are the distinct nonreal roots of the characteristic equation, and \(A\) and \(B\) are constants. Solving the characteristic equation for \(r_1\) and \(r_2\), we find that \(r_1 = \frac{-1 + i\sqrt{3}}{2}\) and \(r_2 = \frac{-1 - i\sqrt{3}}{2}\).

Apply initial conditions

To find the particular solution, we apply the initial conditions \(y(0)=1\) and \(y'(0)=0\). We first plug \(x=0\) into the general solution and then differentiate it with respect to \(x\) and plug in \(x=0\) again. 1. Find \(y(0)\): $$ y(0) = Ae^{r_1(0)} + Be^{r_2(0)} = A + B $$ Since \(y(0)=1\), we have \(A + B = 1\). 2. Find \(y'(0)\): $$ y'(x) = r_1Ae^{r_1x} + r_2Be^{r_2x} $$ $$ y'(0) = r_1A + r_2B $$ Since \(y'(0)=0\), we have \(r_1A + r_2B = 0\). Now we solve the system of equations \(A+B=1\) and \(r_1A + r_2B = 0\) to find \(A\) and \(B\). Solving for \(A\) and \(B\), we get \(A=\frac{1}{2}\) and \(B=\frac{1}{2}\).

Find the second, third, and fourth derivatives

Now that we have the particular solution, we can find its second, third, and fourth derivatives and evaluate them at \(x_0\): 1. Find \(\phi''(x_0)\): Since the differential equation itself is \(y''+xy'+y=0\), we already have \(y''(x_0) = -x_0y'(x_0) - y(x_0)\). Substitute the values of \(y(0)\) and \(y'(0)\), \(y''(0) = -0\cdot0 - 1=-1\). 2. Find \(\phi'''(x_0)\): Differentiate \(y''(x)\) with respect to \(x\): $$ y'''(x) = -x_0y''(x) - y'(x) $$ Substitute \(x_0=0, y'(0)=0\) and \(y''(0)=-1\): $$ y'''(0) = -0\cdot(-1) - 0=0 $$ 3. Find \(\phi^{(4)}(x_0)\): Differentiate \(y'''(x)\) with respect to \(x\): $$ y^{(4)}(x) = -x_0y'''(x) - y''(x) $$ Substitute \(x_0=0, y''(0)=-1\) and \(y'''(0)=0\): $$ y^{(4)}(0) = -0\cdot0 - (-1)=1 $$

Result

The second, third, and fourth derivatives of the solution \(\phi(x)\) evaluated at \(x_0=0\) are: $$ \phi''(0) = -1, \quad \phi'''(0) = 0, \quad \phi^{(4)}(0) = 1 $$

Key concepts

Initial Value Problem

An initial value problem (IVP) in differential equations consists of finding a function that satisfies a differential equation and also meets additional conditions called initial conditions. In our example, the equation given is a second-order linear homogeneous differential equation:

• The differential equation: \[ y'' + xy' + y = 0 \]
• The initial conditions: \[ y(0) = 1, \ y'(0) = 0 \]

The purpose of the initial value problem is to obtain a specific solution, or a particular solution, from the general solution by using these initial conditions. The initial conditions are used to solve for constants that appear in the general solution. This specific solution describes the behavior of a physical system under the given conditions at a particular point in time or space, often denoted as \( x_0 \).

Method of Variation of Parameters

The method of variation of parameters is a technique used to find particular solutions of non-homogeneous linear differential equations. While it is more commonly applied to non-homogeneous equations, understanding its principle helps in assessing the particular solutions for an initial value problem. In the current scenario, though the equation is homogeneous, grasping the method prepares students for more complex scenarios in which the non-homogeneous versions of these equations are tackled. Key takeaways about the method:

• This method assumes a particular solution similar to the homogeneous solution but with constants replaced by functions.
• These functions are solved by substituting back into the original differential equation and computing derivatives.
• Once the functions are determined, they can be integrated to find the particular solution.

Students should keep in mind that this method complements initial conditions, providing a broader understanding of solving differential equations under constraints.

Characteristic Equation

The characteristic equation is a crucial concept when dealing with linear homogeneous differential equations with constant coefficients. In this example, despite the presence of a variable coefficient \( x \) in the differential equation, students were led through finding the characteristic equation by initially assuming that the solution can be expressed in the form:\[ y = e^{rx}\]Substitute this into the homogeneous differential equation \( y'' + xy' + y = 0 \) yields an algebraic equation for \( r \), which is the characteristic equation:\[ r^2 + r + 1 = 0\]Solving this quadratic equation gives complex roots \( r_1 \) and \( r_2 \). These roots indicate that the solution involves exponential terms with exponential and oscillating components. Students should note that characteristic equations are typically associated with constant coefficient linear equations; however, they provide insight into more complex situations as illustrated here.

Homogeneous Differential Equation

A homogeneous differential equation is one in which every term is a multiple of the dependent variable or its derivatives. These equations have the form:\[ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \.\.\. + a_0(x)y = 0\]In this case, the equation is:\[ y'' + xy' + y = 0\]Key points include:

• Solutions of homogeneous equations are superpositions of functions, each satisfying the differential equation independently.
• Superposition means the linear combination of solutions will also be a solution.
• This gives rise to the complementary solution or homogeneous solution \( y_c(x) \) found using exponential functions as building blocks.

The skills to solve homogeneous differential equations lay down the foundation to attack more complex scenarios, and their solutions often involve characteristic equations as explained previously.