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

Short Answer

Expert verified
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

01

Identify the characteristic equation.

For the given 8th order ODE, the characteristic equation is formed by substituting y = e^(r * x) into the ODE: $$ r^8 + 8 r^4 + 16 = 0 $$
02

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: $$ t^2 + 8t + 16 = 0 $$
03

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: $$ t = -4 $$
04

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: $$ r_1 = \sqrt[4]{-4} $$ $$ r_2 = -\sqrt[4]{-4} $$ $$ r_3 = \sqrt[4]{-4}i $$ $$ r_4 = -\sqrt[4]{-4}i $$
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Find the solution of the given initial value problem. Then plot a graph of the solution. \(y^{\prime \prime \prime}+4 y^{\prime}=t, \quad y(0)=y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=1\)

Find a formula involving integrals for a particular solution of the differential equation $$ x^{3} y^{\prime \prime}-3 x^{2} y^{\prime \prime}+6 x y^{\prime}-6 y=g(x), \quad x>0 $$ Hint: Verify that \(x, x^{2},\) and \(x^{3}\) are solutions of the homogenenention.

We consider another way of arriving at the proper form of \(Y(t)\) for use in the method of undetermined coefficients. The procedure is based on the observation that exponential, polynomial, or sinusoidal terms (or sums and products of such terms) can be viewed as solutions of certain linear homogeneous differential equations with constant coefficients. It is convenient to use the symbol \(D\) for \(d / d t\). Then, for example, \(e^{-t}\) is a solution of \((D+1) y=0 ;\) the differential operator \(D+1\) is said to annihilate, or to be an annihilator of, \(e^{-t}\). Similarly, \(D^{2}+4\) is an annihilator of \(\sin 2 t\) or \(\cos 2 t,\) \((D-3)^{2}=D^{2}-6 D+9\) is an annihilator of \(e^{3 t}\) or \(t e^{3 t},\) and so forth. Show that linear differential operators with constant coefficients obey the commutative law, that is, $$ (D-a)(D-b) f=(D-b)(D-a) f $$ for any twice differentiable function \(f\) and any constants \(a\) and \(b .\) The result extends at once to any finite number of factors.

Consider the equation \(y^{\mathrm{iv}}-y=0\) (a) Use Abel's formula [Problem \(20(\mathrm{d}) \text { of Section } 4.1]\) to find the Wronskian of a fundamental set of solutions of the given equation. (b) Determine the Wronskian of the solutions \(e^{t}, e^{-t}, \cos t,\) and \(\sin t .\) (c) Determine the Wronskian of the solutions \(\cosh t, \sinh t, \cos t,\) and \(\sin t .\)

Find the general solution of the given differential equation. Leave your answer in terms of one or more integrals. $$ y^{\prime \prime \prime}-y^{\prime \prime}+y^{\prime}-y=\sec t, \quad-\pi / 2

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