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

Question: Find the general solution of the fourth order linear differential equation: $16y^{(4)}(x) - 8y''(x) + y(x) = 0$ Answer: The general solution of the given differential equation is $y(x) = C_1e^{\frac{1}{2}x} + C_2e^{-\frac{1}{2}x} + C_3xe^{\frac{1}{2}x} + C_4xe^{-\frac{1}{2}x}$, where $C_1$, $C_2$, $C_3$, and $C_4$ are constants determined by the initial conditions.

Step-by-step solution

Assume the solution form

Let's assume a solution of the form \(y(x) = e^{rx}\). We'll find the first, second, and fourth derivatives to substitute into the given differential equation.

Derivative Calculations

Calculate the first, second, and fourth derivatives of the assumed solution: $$ y'(x) = r e^{rx} \\ y''(x) = r^2 e^{rx} \\ y^{(4)}(x) = r^4 e^{rx} $$

Substitute in the Differential Equation

Plugging these derivatives into the given differential equation equation and simplifying, we get: $$ 16(r^4 e^{rx}) - 8(r^2 e^{rx}) + e^{rx} = 0 \\ e^{rx}(16r^4 - 8r^2 + 1) = 0 $$

Solve the Equation for r

Since \(e^{rx}\) is never zero, the equation becomes: $$ 16r^4 - 8r^2 + 1 = 0 $$ Let \(z = r^2\), then we have a quadratic equation with respect to \(z\): $$ 16z^2 - 8z + 1 = 0 $$ Solving the above quadratic equation, we get two values for \(z\): $$ z_1 = \frac{1}{4} \text{ and } z_2 = \frac{1}{4} $$ Since both values are equal, we have only one unique value of z, and therefore we can find the value of r: $$ r = \sqrt{z} = \pm\frac{1}{2} $$

Find the General Solution

Now that we have the values of r (\(\frac{1}{2}\) and \(-\frac{1}{2}\)), we can write the general solution of the given differential equation as: $$ y(x) = C_1e^{\frac{1}{2}x} + C_2e^{-\frac{1}{2}x} + C_3xe^{\frac{1}{2}x} + C_4xe^{-\frac{1}{2}x} $$ where \(C_1\), \(C_2\), \(C_3\), and \(C_4\) are constants determined by the initial conditions.

Key concepts

General Solution

The general solution of a differential equation is a crucial concept that signifies a family of functions satisfying the equation. For an equation like \[ 16 y^{(4)} - 8 y'' + y = 0 \], finding the general solution involves a systematic approach.

• Assume a trial solution, often using terms of the form \(y(x) = e^{rx}\), to explore the possible outcomes of the differential equation.
• Find the derivatives of the assumed solution as required by the equation.
• Substitute these derivatives back into the original equation to form an algebraic equation.
• Solving this will lead to a characteristic equation, whose roots help determine the nature of the solution.

In this particular exercise, after calculating and simplifying, the general solution is established as a combination of terms with exponential functions: \[ y(x) = C_1e^{\frac{1}{2}x} + C_2e^{-\frac{1}{2}x} + C_3xe^{\frac{1}{2}x} + C_4xe^{-\frac{1}{2}x} \].The constants \(C_1\), \(C_2\), \(C_3\), and \(C_4\) are parameters that are later determined when additional information, like initial conditions, is provided.

Initial Value Problem

An initial value problem (IVP) gives specific conditions at the initial point of a function's domain. This concept is essential because it transforms a general solution into a particular one that fits the criteria of a real-world scenario.

• This involves not only solving the differential equation but also using initial conditions to find the particular constants in the general solution.
• The main goal is to provide a unique solution utilizing the intertwined nature of initial data and the differential relations.

If an initial condition were provided right after determining the general solution from the previous section, such as \(y(0) = y_0\) and possibly derivatives like \(y'(0) = y_1\), one would substitute these conditions into the general solution. The aim is to solve for the constants \(C_1, C_2, C_3,\) and \(C_4\) to obtain a specific function that uniquely satisfies both the differential equation and the initial conditions.

Homogeneous Differential Equation

A homogeneous differential equation has the right-hand side set to zero, reflecting a balance or symmetry in the system being modeled. In mathematical terms, an equation of the form \[ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \, ... \, + a_1 y' + a_0 y = 0 \] is homogeneous if all terms involve the dependent variable or its derivatives.

• For this exercise, the differential equation \(16 y^{(4)} - 8 y'' + y = 0\) is homogeneous since all terms on the left combine to form zero.
• Such equations often describe natural phenomena or systems in equilibrium.
• Finding solutions involves looking for linear combinations of certain fundamental functions that inherently satisfy the equation.

By assuming a solution form, as we did with \(e^{rx}\), you can eventually express the solution as a superposition of these fundamental functions, reflective of different system modes or states.

Characteristic Equation

Characteristic equations are pivotal tools in solving linear differential equations, particularly when dealing with constant coefficients. By translating a differential equation into an algebraic form, it becomes feasible to ascertain the type and form of its solutions.

• For the given differential equation \(16 y^{(4)} - 8 y'' + y = 0\), substituting the assumed exponential function leads to an expression like \(e^{rx}(16r^4 - 8r^2 + 1) = 0\).
• Since \(e^{rx}\) is never zero, solve for the polynomial \(16r^4 - 8r^2 + 1 = 0\) to find the roots \(r\), which characterizes the solution.
• The characteristic equation is typically quadratic or of higher order, and the nature of its roots (real, complex, or repeated) determines the form of the general solution.
• In this case, solving the characteristic equation gives \(z = \frac{1}{4}\) repeatedly, reflecting repeated real roots \(-\frac{1}{2}\) and \(\frac{1}{2}\).

These roots guide the structure of the solution to the differential equation, leading, in repeated root instances, to terms involving both exponentials and polynomials, as seen in the general solution provided earlier.