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Chapter 4: Problem 21

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

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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.

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

Follow the procedure illustrated in Example 4 to determine the indicated roots of the given complex number. $$ (1-i)^{1 / 2} $$

Find the general solution of the given differential equation. $$ y^{\text {iv }}-7 y^{\prime \prime \prime}+6 y^{\prime \prime}+30 y^{\prime}-36 y=0 $$

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. Consider the problem of finding the form of the particular solution \(Y(t)\) of $$ (D-2)^{3}(D+1) Y=3 e^{2 t}-t e^{-t} $$ where the left side of the equation is written in a form corresponding to the factorization of the characteristic polynomial. (a) Show that \(D-2\) and \((D+1)^{2},\) respectively, are annihilators of the terms on the right side of Eq. and that the combined operator \((D-2)(D+1)^{2}\) annihilates both terms on the right side of Eq. (i) simultaneously. (b) Apply the operator \((D-2)(D+1)^{2}\) to Eq. (i) and use the result of Problem 20 to obtain $$ (D-2)^{4}(D+1)^{3} Y=0 $$ Thus \(Y\) is a solution of the homogeneous equation (ii). By solving Eq. (ii), show that $$ \begin{aligned} Y(t)=c_{1} e^{2 t} &+c_{2} t e^{2 t}+c_{3} t^{2} e^{2 t}+c_{4} t^{3} e^{2 t} \\ &+c_{5} e^{-t}+c_{6} t e^{-t}+c_{7} t^{2} e^{-t} \end{aligned} $$ where \(c_{1}, \ldots, c_{7}\) are constants, as yet undetermined. (c) Observe that \(e^{2 t}, t e^{2 t}, t^{2} e^{2 t},\) and \(e^{-t}\) are solutions of the homogeneous equation corresponding to Eq. (i); hence these terms are not useful in solving the nonhomogeneous equation. Therefore, choose \(c_{1}, c_{2}, c_{3},\) and \(c_{5}\) to be zero in Eq. (iii), so that $$ Y(t)=c_{4} t^{3} e^{2 t}+c_{6} t e^{-t}+c_{7} t^{2} e^{-t} $$ This is the form of the particular solution \(Y\) of Eq. (i). The values of the coefficients \(c_{4}, c_{6},\) and \(c_{7}\) can be found by substituting from Eq. (iv) in the differential equation (i).

Find the solution of the given initial value problem. Then plot a graph of the solution. \(y^{\mathrm{iv}}+2 y^{\prime \prime \prime}+y^{\prime \prime}+8 y^{\prime}-12 y=12 \sin t-e^{-t}, \quad y(0)=3, \quad y^{\prime}(0)=0\) \(y^{\prime \prime}(0)=-1, \quad y^{\prime \prime \prime}(0)=2\)

Determine the general solution of the given differential equation. \(y^{\mathrm{iv}}-4 y^{\prime \prime}=t^{2}+e^{t}\)
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