Question

Find the general solution of the given differential equation. $$ y^{\text {viii }}+8 y^{\text {iv }}+16 y=0 $$

Short answer

Answer: The general solution for the given differential equation is: $$ y(x) = C_1 e^{\sqrt[4]{-4} x} + C_2 e^{-\sqrt[4]{-4} x} + C_3 e^{\sqrt[4]{-4} xi} + C_4 e^{-\sqrt[4]{-4} xi} $$

Step-by-step solution

Identify the characteristic equation.

For the given 8th order ODE, the characteristic equation is formed by substituting y = e^(r * x) into the ODE: $$ r^8 + 8 r^4 + 16 = 0 $$

Solve the characteristic equation.

To solve the characteristic equation \(r^8 + 8 r^4 + 16 = 0\), let's make a substitution. Let \(t = r^4\), then our equation becomes a quadratic equation: $$ t^2 + 8t + 16 = 0 $$

Find the roots of the quadratic equation.

Now, we will find the roots of the quadratic equation \(t^2 + 8t + 16 = 0\). To do this, we can use the quadratic formula: $$ t = \frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ where a=1, b=8, and c=16. Plugging in the values, we get: $$ t = \frac{-8\pm\sqrt{8^2-4(16)}}{2(1)} $$ Since the discriminant (\(b^2 - 4ac = 64 - 64 = 0\)) is 0, we have a repeated root: $$ t = -4 $$

Find the roots of the original characteristic equation.

Now that we found the root of the quadratic equation, \(t=-4\), we need to find the roots of the original characteristic equation by finding the values of r that satisfy \(t=r^4\). Since we have only one value for t, we will have 4 distinct roots for r: $$ r = \pm\sqrt[4]{-4}, \pm \sqrt[4]{-4}i $$ So the roots are: $$ r_1 = \sqrt[4]{-4} $$ $$ r_2 = -\sqrt[4]{-4} $$ $$ r_3 = \sqrt[4]{-4}i $$ $$ r_4 = -\sqrt[4]{-4}i $$

Write the general solution of the given ODE.

Now that we have found the roots of our characteristic equation, we can write the general solution of our given 8th-order differential equation using the form: $$ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} + C_3 e^{r_3 x} + C_4 e^{r_4 x} $$ Substituting our roots, we get the general solution: $$ y(x) = C_1 e^{\sqrt[4]{-4} x} + C_2 e^{-\sqrt[4]{-4} x} + C_3 e^{\sqrt[4]{-4} xi} + C_4 e^{-\sqrt[4]{-4} xi} $$ This is the general solution for the given differential equation.

Key concepts

Characteristic Equation

When solving a differential equation, the characteristic equation is a crucial step in finding the solution. It is derived from the original differential equation by assuming a solution of the form \( y = e^{rx} \), where \( r \) is a constant. This substitution simplifies the differential equation into an algebraic equation in terms of \( r \).

In the given example with the 8th order ODE \( y^{\text{viii}} + 8y^{\text{iv}} + 16y = 0 \), the characteristic equation becomes \( r^8 + 8r^4 + 16 = 0 \). Solving this algebraic equation will yield the values of \( r \) that are then used to construct the general solution. Understanding how to form and solve the characteristic equation is fundamental in solving higher-order linear differential equations with constant coefficients.

General Solution

The general solution of a differential equation represents the complete set of all possible solutions that satisfy the equation. It incorporates arbitrary constants \( C_1, C_2, \ldots \), which are determined by initial conditions or boundary values.

For linear differential equations with constant coefficients, once the characteristic equation is solved and the roots are obtained, these roots are used to form the general solution. If the differential equation is of the nth order, there will be n terms in the general solution, each term associated with a root of the characteristic equation and a corresponding constant \( C_i \). The general solution of our example is a linear combination of exponentials of the roots, ensuring that it encompasses all possible solutions for the given ODE.

Quadratic Formula

The quadratic formula is a staple in algebra that provides the solutions to quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a \), \( b \), and \( c \) are coefficients of the equation and the discriminant \( b^2 - 4ac \) determines the nature of the roots.

In our step-by-step solution, the quadratic formula is used to find the roots of the reduced characteristic equation, \( t^2 + 8t + 16 = 0 \). Knowing how to apply this formula correctly is essential for solving the characteristic equation when it can be reduced to a quadratic form.

Complex Roots

When the discriminant of the quadratic equation \( b^2 - 4ac \) is negative, the equation has complex roots, which are in the form of \( a \pm bi \), where \( i \) is the imaginary unit and \( a \), \( b \) are real numbers.

In the context of differential equations, complex roots occur when finding the values of \( r \) in the characteristic equation. Once complex roots are identified, they contribute to the general solution in sinusoidal terms, reflecting the oscillating components of the solution. In our exercise, the roots include \( \sqrt[4]{-4} \) and its multiples, which give rise to complex numbers when the fourth root is applied. It's important to understand how complex roots translate into the terms of the general solution and how they indicate the presence of oscillatory behavior in the system described by the differential equation.