Question

In each exercise, (a) Find the general solution of the differential equation. (b) If initial conditions are specified, solve the initial value problem. $$ 16 y^{(4)}+8 y^{\prime \prime}+y=0 $$

Short answer

The general solution of the given fourth-order linear homogeneous differential equation is: $$ y(t) = A\cos(\frac{1}{2}t) + B\sin(\frac{1}{2}t) $$ where A and B are constants that can be determined using the initial conditions if they were provided.

Step-by-step solution

Create the characteristic equation

Given the differential equation, we can form the characteristic equation by substituting each derivative term \(y^{(n)}\) with \(r^n\). The characteristic equation will be: $$ 16r^4 + 8r^2 + 1 = 0 $$

Find the roots of the characteristic equation

To find the roots of the characteristic equation, we can make a substitution for better visualization. Let \(x = r^2\). So, our equation becomes: $$ 16x^2 + 8x + 1 = 0 $$ This is a quadratic equation in form of \(ax^2+bx+c = 0\), and we can solve for \(x\) using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Substitute \(a = 16\), \(b= 8\), and \(c = 1\): $$ x = \frac{-8 \pm \sqrt{8^2 - 4(16)(1)}}{2(16)} $$ By evaluating the expression above, we obtain: $$ x = -\frac{8 \pm \sqrt{64 - 64}}{32} $$ $$ x = -\frac{8 \pm 0}{32} $$ $$ x = -\frac{1}{4} $$ Now we need to go back to the original variable \(r\). Since \(x = r^2\), we have: $$ r^2 = -\frac{1}{4} $$ Taking the square root of both sides, we get \(r = \pm\frac{1}{2}i\). So, we have two complex conjugate roots: $$ r_1 = \frac{1}{2}i, r_2 = -\frac{1}{2}i $$

Construct the general solution

Now that we have the roots of the characteristic equation, we can construct the general solution to the given differential equation. Since we have complex conjugate roots, the general solution will take the form: $$ y(t) = C_1 e^{-\frac{1}{2}it} + C_2 e^{\frac{1}{2}it} $$ However, since dealing with complex exponential functions is not very practical, we can express the general solution in terms of sine and cosine, using Euler's formula \(e^{ix}=\cos(x)+i\sin(x)\): $$ y(t) = C_1 (\cos(\frac{1}{2}t) + i\sin(\frac{1}{2}t)) + C_2 (\cos(-\frac{1}{2}t) + i\sin(-\frac{1}{2}t)) $$ By factoring out the real and imaginary parts, we can rewrite the general solution as: $$ y(t) = (C_1 + C_2)\cos(\frac{1}{2}t) + (iC_1 - iC_2)\sin(\frac{1}{2}t) $$ Finally, let \(A = C_1 + C_2\) and \(B = iC_1 - iC_2\), then the general solution is: $$ y(t) = A\cos(\frac{1}{2}t) + B\sin(\frac{1}{2}t) $$ As there are no initial conditions specified in the exercise, we do not need to solve the initial value problem.

Key concepts

Characteristic Equation

Understanding the characteristic equation is crucial when it comes to solving linear differential equations with constant coefficients. It's a bridge between the differential equation and the algebraic domain, where the differential operators are replaced with polynomial terms that are much easier to handle.

In the given exercise, the differential equation is expressed as a fourth-order linear homogeneous equation with constant coefficients. To solve such an equation, we replace each derivative of the function with powers of an unknown variable, typically denoted as r, to form the characteristic equation. The general form of the characteristic equation for a fourth-order linear differential equation would be:
\[ a r^4 + b r^3 + c r^2 + d r + e = 0 \],
where a, b, c, d, and e are constants that correspond to the coefficients in the original differential equation.

For our specific problem, we get the characteristic equation:\[ 16r^4 + 8r^2 + 1 = 0 \],
which simplifies the problem to finding roots of a polynomial, a task for which we have a well-developed set of tools in algebra. The roots of this equation will determine the form of the general solution to the original differential equation.

Initial Value Problem

An initial value problem in the context of differential equations involves finding a specific solution that not only satisfies the differential equation but also meets initial conditions at a given point. These initial conditions provide us with necessary information to determine the constants in the general solution that would typically represent an infinite set of possible solutions.

In our example, the differential equation didn't come with initial conditions, meaning that we're left with the task of finding just the general solution. However, if initial conditions, such as y(0) for the function's value and y'(0) for the first derivative's value at t=0, had been given, we would proceed by plugging these into the general solution derived from the characteristic equation to find the exact values of the constants involved. This added step refines the general solution to one specific solution that traces a unique path through the point defined by the initial conditions.

General Solution of Differential Equation

The general solution of a differential equation is a formula that includes all possible solutions to the equation. It incorporates arbitrary constants, usually denoted as C₁, C₂, and so on, which represent the variety of solutions that satisfy the differential equation without any additional constraints.

For the differential equation presented in the exercise, the general solution is a combination of sine and cosine functions with coefficients that will be determined by initial conditions, if any are provided. The sine and cosine functions arise when dealing with complex roots of the characteristic equation, as Euler's formula connects complex exponentials to these trigonometric functions.

The final form of the general solution for our differential equation, as given in the step-by-step solution, is:\[ y(t) = A\text{cos}(\frac{1}{2}t) + B\text{sin}(\frac{1}{2}t) \],
where A and B are constants derived from the initial conditions or, in the absence thereof, remain arbitrary constants that can be tailored to fit particular scenarios imposed by specific conditions. This elegant expression encapsulates all possible behaviors of the system described by the original differential equation.