Question

Solve the following initial value problems. (a) \(\frac{d^{4} y}{d x^{4}}=y, \quad y(0)=y^{\prime}(0)=y^{\prime \prime \prime}(0)=0, \quad y^{\prime \prime}(0)=1\), (b) \(\frac{d^{4} y}{d x^{4}}+\frac{d^{2} y}{d x^{2}}=0, \quad y(0)=y^{\prime \prime}(0)=y^{\prime \prime \prime}(0)=0, \quad y^{\prime}(0)=1\), (c) \(\frac{d^{4} y}{d x^{4}}=0, \quad y(0)=y^{\prime}(0)=y^{\prime \prime}(0)=0, \quad y^{\prime \prime \prime}(0)=2\).

Short answer

The solutions for the problems are (a) \(y(x) = -\cos(x)\), (b) \(y(x) = -\frac{1}{2}\cos(x) + \sin(x) + \frac{1}{2}e^x\), (c) \(y(x) = \frac{1}{6} x^{3}\).

Step-by-step solution

Problem (a) - Obtain the general solution

First, we solve the differential equation \( \frac{d^{4} y}{d x^{4}}=y \) using the characteristic equation. The characteristic equation is \(m^{4} - 1 = 0\), whose roots are \(m = \pm 1, \pm i\). This gives the general solution as \(y(x) = c_1 e^x + c_2 e^{-x} + c_3 \cos(x) + c_4 \sin(x)\).

Problem (a) - Apply the initial value conditions

Apply the initial values \(y(0)=y^{\prime}(0)=y^{\prime \prime \prime}(0)=0, y^{\prime \prime}(0)=1\) to the derived general solution to get \(c_1 + c_2 = 0\), \(c_1 - c_2 + c_4 = 0\), \(c_1 - c_2 - c_3 = 0\) and \(c_1 + c_2 + c_3 = 0\). Solving these simultaneous equations gives us \(c_1 = 0, c_2 = 0, c_3 = -1, c_4 = 0\). Therefore, the solution to problem (a) is \(y(x) = -\cos(x)\).

Problem (b) - Obtain the general solution

The equation \( \frac{d^{4} y}{d x^{4}}+\frac{d^{2} y}{d x^{2}}=0 \) characteristic equation is \(m^{4} + 1 = 0\), whose roots are \(m = \pm i, \pm 1\). Therefore, the general solution is given by \(y(x) = c_1 \cos(x) + c_2 \sin(x) + c_3 e^x + c_4 e^{-x}\).

Problem (b) - Apply the initial value conditions

We then apply the initial values \(y(0)=y^{\prime \prime}(0)=y^{\prime \prime \prime}(0)=0, y^{\prime}(0)=1\). This gives us \(c_1 + c_3 + c_4 = 0\), \(c_2 + c_3 - c_4 = 1\), \(c_1 - c_2 + c_3 - c_4 = 0\) and \(c_2 - c_1 + c_3 - c_4 = 0\). Solving this set of simultaneous equations will give us \(c_1 = -\frac{1}{2}, c_2 = 1, c_3 = \frac{1}{2}, c_4 = 0\). Therefore, the solution to problem (b) is \(y(x) = -\frac{1}{2}\cos(x) + \sin(x) + \frac{1}{2}e^x\).

Problem (c) - Obtain the general solution

The equation \( \frac{d^{4} y}{d x^{4}} = 0 \) has a general solution of \(y(x) = c_1 + c_2 x + c_3 x^{2} + c_4 x^{3}\).

Problem (c) - Apply the initial value conditions

By applying the initial values \(y(0)=y^{\prime}(0)=y^{\prime \prime}(0)=0, y^{\prime \prime \prime}(0)=2\), we obtain \(c_1 = 0\), \(c_2 = 0\), \(c_3 = 0\) and \(c_4 = \frac{1}{6}\). Thus, the solution to problem (c) is \(y(x) = \frac{1}{6} x^{3}\).

Key concepts

Characteristic Equation

When solving differential equations, the characteristic equation plays a crucial role, especially for constant coefficient linear differential equations. It transforms a differential equation into an algebraic equation, making it easier to find solutions.

• For equation (a), the differential equation is \( \frac{d^{4} y}{d x^{4}}=y \). The characteristic equation derived from this is \( m^{4} - 1 = 0 \).
• This characteristic equation is solved for \( m \), resulting in roots \( m = \pm 1, \pm i \).
• The general solution involves exponential terms like \( e^{x} \) and trigonometric functions like \( \cos(x) \), based on these roots.

Understanding how to derive and solve the characteristic equation helps in constructing the general solution for differential equations, which is the first step before applying specific conditions.

Initial Value Problem

An initial value problem specifies the values of the function and its derivatives at a certain point, usually at \( x = 0 \). These conditions help us find the specific solution to a differential equation.

• In problem (a), the initial values are \( y(0) = 0 \), \( y'(0) = 0 \), \( y''(0) = 1 \), and \( y'''(0) = 0 \).
• These values are applied to the general solution obtained after solving the characteristic equation.
• By substituting these values into the general solution, a system of equations is formed. Solving this system gives the specific constants \( c_1, c_2, c_3, \) and \( c_4 \) for the unique solution.

Solving an initial value problem is essential to pinpoint the exact behavior of a system described by a differential equation under given conditions.

Higher Order Derivatives

Differential equations often involve higher order derivatives, which are derivatives of a function taken multiple times. Understanding how to manage and solve these equations is crucial.

• In these exercises, fourth-order derivatives \( \frac{d^{4} y}{dx^{4}} \) are featured prominently, representing complex systems or phenomena.
• Higher order derivatives translate into polynomial characteristic equations, which can then be solved for roots.
• Once the characteristic equation is solved, it gives insights into the possible behaviors of the system, like oscillations or exponential growth/decay.

Mastering higher order derivatives and their corresponding differential equations allows one to tackle advanced mathematical problems and model real-world systems effectively.