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. $$ y^{\prime \prime}+(\sin x) y^{\prime}+(\cos x) y=0 ; \quad y(0)=0, \quad y^{\prime}(0)=1 $$

Short answer

$$\begin{aligned} y^{\prime\prime}+(\sin x)y^{\prime}+(\cos x)y &= 0 \\ y(0) &= 0 \\ y^{\prime}(0) &= 1 \end{aligned}$$ Answer: \(\phi^{\prime\prime}(0) = -1\), \(\phi^{\prime\prime\prime}(0) = 2\), and \(\phi^{\mathrm{iv}}(0) = -1\).

Step-by-step solution

Write down the given information

We are given the following initial value problem: $$\begin{aligned} y^{\prime\prime}+(\sin x)y^{\prime}+(\cos x)y &= 0 \\ y(0) &= 0 \\ y^{\prime}(0) &= 1 \end{aligned}$$ Our task is to find \(\phi^{\prime\prime}(x_0)\), \(\phi^{\prime\prime\prime}(x_0)\), and \(\phi^{\mathrm{iv}}(x_0)\), for the given point \(x_{0}\) if \(y=\phi(x)\) is a solution of the given initial value problem. Our starting point is the given second-order differential equation: $$y^{\prime\prime}+(\sin x)y^{\prime}+(\cos x)y = 0$$

Differentiate the given equation with respect to x

To find the higher-order derivatives, we will differentiate the given differential equation with respect to \(x\). We obtain the third-order differential equation: $$y^{\prime\prime\prime}+(\sin x)y^{\prime\prime}+(\cos x)y^{\prime}- (\sin x)y^{\prime}+(\cos x)y^{\prime\prime}-(\sin x)y = 0$$ Simplifying and rearranging the terms, we get: $$y^{\prime\prime\prime}+(\sin x + \cos x)y^{\prime\prime}+(\cos x - \sin x)y^{\prime}-(\sin x)y = 0$$ Now we will differentiate this third-order differential equation one more time to find the fourth-order derivative \(\phi^{\mathrm{iv}}(x)\):

Differentiate the third-order equation with respect to x

$$y^{\mathrm{iv}}+(\sin x + \cos x)y^{\prime\prime\prime}+(\cos x - \sin x)y^{\prime\prime}-(\sin x + \cos x)y^{\prime}-(\cos x)y = 0$$ Now we will use the initial conditions \(y(0) = 0, y^{\prime}(0) = 1\) to find the value of the higher-order derivatives at \(x_0 = 0\).

Apply initial conditions and find higher-order derivatives at x = 0

In the given second-order differential equation, substituting \(x = 0\) and applying initial conditions \(y(0) = 0, y^{\prime}(0) = 1\), we obtain: $$y^{\prime\prime}(0)+1=0$$ Hence, \(y^{\prime\prime}(0) = -1\). Now, we substitute \(x = 0\) and \(y^{\prime\prime}(0) = -1\) in the third-order differential equation: $$y^{\prime\prime\prime}(0)-1 -1 = 0$$ Hence, \(y^{\prime\prime\prime}(0) = 2\). Finally, we substitute \(x = 0\), \(y^{\prime\prime}(0) = -1\), and \(y^{\prime\prime\prime}(0) = 2\) in the fourth-order differential equation: $$y^{\mathrm{iv}}(0) + 2 -1 -1 + 1 = 0$$ Hence, \(y^{\mathrm{iv}}(0) = -1\).

Write down the obtained higher-order derivatives values at x = 0

We found: $$\begin{aligned} \phi^{\prime\prime}(0) &=-1 \\ \phi^{\prime\prime\prime}(0) &= 2 \\ \phi^{\mathrm{iv}}(0) &=-1 \end{aligned}$$ So, the higher-order derivatives of the function \(\phi(x)\) at \(x_0 = 0\) are \(\phi^{\prime\prime}(0) = -1\), \(\phi^{\prime\prime\prime}(0) = 2\), and \(\phi^{\mathrm{iv}}(0) = -1\).

Key concepts

Differential Equations

Differential equations are powerful tools in mathematics and the sciences, representing relationships between functions and their derivatives. They tell us how a particular quantity changes with respect to another, often over time or space.

When we talk about solving a differential equation, we seek a function or a set of functions that satisfy the equation. An initial value problem is a special type of differential equation where we are provided with the value of the function, and possibly some of its derivatives, at a specific point. In the given exercise, we are asked to solve a second-order differential equation with initial values provided at the point zero.

The process begins by understanding the presented equation and initial conditions, which are critical for determining the behavior of the function at the starting point. By applying the given initial conditions, we can integrate step by step, each time considering the next higher-order derivative, until we find the function values required.

Higher-Order Derivatives

Higher-order derivatives are simply the derivatives of a function taken multiple times. For instance, if we take the derivative of a function once, we get the first derivative, often denoted as \( f'(x) \). Taking the derivative of \( f'(x) \) gets us the second derivative, \( f''(x) \) or \( \frac{d^{2}}{dx^{2}}f(x) \), and so on.

The significance of higher-order derivatives lies in their ability to capture more sophisticated changes in a function's behavior, such as acceleration in physical contexts. When solving the initial value problem in the example, we differentiated the original second-order differential equation twice more to obtain the third and fourth derivatives. The resulting equations were then used along with the initial conditions to find the specific values of these derivatives at the point \( x_0 \).

It is crucial to interpret the initial conditions correctly and apply them methodically at each differentiation stage. This process ensures that we account for the specific conditions of the function as we peel back its layers of complexity through each higher-order derivative.

Second-Order Differential Equation

Second-order differential equations involve the second derivative of a function and can describe a wide range of phenomena, including oscillations and resonance in mechanical systems, or the curvature of a graph. These equations may look daunting at first, but they follow systematic solving procedures that can be learned with practice.

The given exercise featured a second-order differential equation with sinusoidal coefficients and initial conditions, a classic problem type in differential equations. In real-world scenarios, such equations often model physical processes where the rate of change itself is changing, like the motion of a pendulum or an electrical circuit's behavior.

Solving such an equation involves finding a general solution that would work for any initial condition and then applying the specific initial conditions provided. This process leads us to the particular solution applicable to the scenario in question. Our exercise demonstrated this by using the given initial values to calculate specific higher-order derivatives' values, which are critical for predicting the function's behavior beyond the initial point.