Theory of Higher-Order Linear Differential Equations

Solve the given initial value problem 

 $\begin{array}{l}y\text{'}\text{'}\text{'}+7y\text{'}\text{'}+14y\text{'}+8y=0\\ y\left(0\right)=1\\ y\text{'}\left(0\right)=-3\\ y\text{'}\text{'}\left(0\right)=13\end{array}$

Solve the given initial value problem

_{$\begin{array}{l}y\text{'}\text{'}\text{'}-4y\text{'}\text{'}+7y\text{'}-6y=0\\ y\left(0\right)=1\\ y\text{'}\left(0\right)=0\\ y\text{'}\text{'}\left(0\right)=0\end{array}$}

Find a general solution for the given linear system using the elimination method of Section 5.2.

$\begin{array}{l}\frac{d^{2}x}{d{t}^2}-x+5y=0\\ 2x+\frac{d^{2}y}{d{t}^2}+2y=0\end{array}$

Find a general solution for the given

linear system using the elimination method of Section 5.2.

$\begin{array}{l}\frac{d^{3}x}{d{t}^3}-x+\frac{dy}{dt}+y=0\\ \frac{dx}{dt}-x+y=0\end{array}$

Let $P\left(r_1\right)=a_n{r}_1^n+a_{n-1}{r}_1^{n-1}+...+a_1{r}_1+a_0$  be a polynomialwith real coefficients $a_n,....,a_0$  . Prove that if r₁ is azero of  $P\left(r\right)$ , then so is its complex conjugate r₁. [Hint:Show that $\overline{P\left(r\right)}=P\left(\overline{r}\right)$  , where the bar denotes complexconjugation.]

Show that the m functions   are linearly dependent on (-∞,∞) [Hint: Show thatthese functions are linearly independent if and only if1, x, . . . x^m-1, are linearly independent.] 

As an alternative proof that the functions  $e^{r_{1}x},e^{r_{2}x},e^{r_{3}x},\mathrm{...},e^{r_{n}x}$ are linearly independent on (∞,-∞) when $r_1,r_2,...r_n$  are distinct, assume $C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}+C_3{e}^{r_{3}x}+...+C_n{e}^{r_{n}x}$ holds for all x in (∞,-∞)  and proceed as follows:

(a) Because the ri’s are distinct we can (if necessary)relabel them so that $r_1>r_2>r_3>...>r_n$.Divide equation (33) by  to obtain $C_1+\frac{C_2{e}^{r_{2}x}}{e^{r_{2}x}}+\frac{C_3{e}^{r_{3}x}}{e^{r_{3}x}}+...+\frac{C_n{e}^{r_{n}x}}{e^{r_{n}x}}=0$ Now let x→∞ on the left-hand side to obtain C1 = 0. (b) Since C1 = 0, equation (33) becomes

 $C_2{e}^{r_{2}x}+C_3{e}^{r_{3}x}+...+C_n{e}^{r_{n}x}$ = 0for all x in(∞,-∞). Divide this equation by  $e^{r_{2}x}$

and let x→∞ to conclude that C2 = 0.

(c) Continuing in the manner of (b), argue that all thecoefficients, C1, C2, . . . ,Cn are zero and hence $e^{r_{1}x},e^{r_{2}x},e^{r_{3}x},...,e^{r_{n}x}$are linearly independent on (∞,-∞).

Find a general solution to

 $y^{\left(4\right)}+2y\text{'}\text{'}\text{'}-3y\text{'}\text{'}-y\text{'}+\frac{1}{2}y=0$

by using Newton’s method (Appendix B) or some othernumerical procedure to approximate the roots of the auxiliaryequation.

Find a general solution to y’’’ - 3y’ - y = 0 by using Newton’s method or some other numerical procedure to approximate the roots of the auxiliary equation.

Find a general solution to  $y^{\left(4\right)}+2y^{\left(3\right)}+4y\text{'}\text{'}+3y\text{'}+2y=0$ by using Newton’s method to approximate numerically the roots of the auxiliary equation. [Hint: To find complex roots, use the Newton recursion formula $z_{n+1}=z_n-\frac{f\left(z_n\right)}{f\text{'}\left(z_n\right)}$ and start with a complex initial guess z0.]

Higher-Order Cauchy–Euler Equations. A differential equation that can be expressed in the form

 $a_n{x}^n{y}^{\left(n\right)}\left(x\right)+a_{n-1}{x}^{n-1}{y}^{(n-1)}\left(x\right)+...+a_{0}y\left(x\right)=0$

where $a_n,a_{n-1},....,a_0$ are constants, is called a homogeneous Cauchy–Euler equation. (The second-order case is discussed in Section 4.7.) Use the substitution  $y=x^y$ to help determine a fundamental solution set for the following Cauchy–Euler equations:

(a)  $x^{3}y\text{'}\text{'}\text{'}+x^{2}y\text{'}\text{'}-2xy\text{'}+2y=0,x>0$

(b)  $x^4{y}^{\left(4\right)}+6x^{3}y\text{'}\text{'}\text{'}+2x^{2}y\text{'}\text{'}-4xy+4y=0,x>0$

(c)  $x^{3}y\text{'}\text{'}\text{'}-2x^{2}y\text{'}\text{'}+13xy\text{'}-13y=0,x>0$

[Hint:   $x^{\alpha +\beta i}=e^{(\alpha +\beta i)lnx}=x^a\left\{cos\right(\beta lnx)+isin(\beta lnx\left)\right\}$]

(a) Derive the form  $y\left(x\right)=A_1{e}^x+A_2{e}^{-x}+A_{3}cosx+A_{4}sinx$ for the general solution to the equation  $y^{\left(4\right)}=y$, from the observation that the fourth roots of unity are 1, -1, i, and -i. 

(b) Derive the form 

 $y\left(x\right)=A_1{e}^x+A_2{e}^{-x/2}cos\left(\frac{\sqrt{3x}}{2}\right)+A_3{e}^{-x/2}sin\left(\frac{\sqrt{3x}}{2}\right)$

for the general solution to the equation  $y^{\left(3\right)}=y$ from the observation that the cube roots of unity are 1, $e^{i\frac{2\pi }{3}}$  , and  $e^{-i\frac{2\pi }{3}}$.

Let y₁x₂= Ce^rx, where C (≠0) and r are real numbers,be a solution to a differential equation. Supposewe cannot determine r exactly but can only approximateit by $r$ . Let   (x) =Ce^rxand consider the error

 $\left|y\left(x\right)-\tilde{y}\left(x\right)\right|$

(a) If r and  $\tilde{r}$ are positive, r ≠­ , show that the errorgrows exponentially large as x approaches + ∞.

(b) If r and  $r$ are negative, r≠ , show that the errorgoes to zero exponentially as x approaches + ∞.

On a smooth horizontal surface, a mass of m₁ kg isattached to a fixed wall by a spring with spring constantk₁ N/m. Another mass of m₂ kg is attached to thefirst object by a spring with spring constant k₂ N/m. Theobjects are aligned horizontally so that the springs aretheir natural lengths. As we showed in Section 5.6, thiscoupled mass–spring system is governed by the systemof differential equations

 $m_1\frac{d^{2}x}{d{t}^2}+(k_1+k_2)x-k_{2}y=0$

 $m_2\frac{d^{2}y}{d{t}^2}-k_{2}x+k_{2}y=0$

Let’s assume that m₁ = m₂ = 1, k₁ = 3, and k₂ = 2.If both objects are displaced 1 m to the right of theirequilibrium positions (compare Figure 5.26, page 283)and then released, determine the equations of motion forthe objects as follows:

(a)Show that x(2) satisfies the equation $x^{\left(4\right)}\left(t\right)+7x\text{'}\text{'}\left(t\right)+6x\left(t\right)=0$

(b) Find a general solution x(2) to (36).

(c) Substitute x(2) back into (34) to obtain a generalsolution for y(2)

(d) Use the initial conditions to determine the solutions,x(2) and y(2), which are the equations of motion.

Suppose the two springs in the coupled mass–spring system discussed in Problem 33 are switched, giving the new data \({\bf{m1}}{\rm{ }} = {\rm{ }}{\bf{m2}}{\rm{ }} = {\rm{ }}{\bf{1}},{\rm{ }}{\bf{k1}}{\rm{ }} = {\rm{ }}{\bf{2}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{k2}}{\rm{ }} = {\rm{ }}{\bf{3}}\)  If both objects are now displaced 1 m to the right of their equilibrium positions and then released, determine the equations of motion of the two objects.

use the method of undetermined coefficients to determine the form of a particular solution for the given equation.

$y\text{'}\text{'}\text{'}-2y\text{'}\text{'}-5y\text{'}+6y=e^x+x^2$

use the method of undetermined coefficients to determine the form of a particular solution for the given equation.

$y\text{'}\text{'}\text{'}+y\text{'}\text{'}-5y\text{'}+3y=e^{-x}+sinx$

use the method of undetermined coefficients to determine the form of a particular solution for the given equation.

$y\text{'}\text{'}\text{'}+3y\text{'}\text{'}-4y=e^{-2x}$

use the method of undetermined coefficients to determine the form of a particular solution for the given equation.

$y\text{'}\text{'}\text{'}+y\text{'}\text{'}-2y=x{e}^x+1$

find a general solution to the given equation.

$y\text{'}\text{'}\text{'}-2y\text{'}\text{'}-5y\text{'}+6y=e^x+x^2$

find a general solution to the given equation. $y\text{'}\text{'}\text{'}+y\text{'}\text{'}-5y\text{'}+3y=e^{-x}+sinx.$

find a general solution to the given equation.

$y\text{'}\text{'}\text{'}+3y\text{'}\text{'}-4y=e^{-2x}$

find a general solution to the given equation.

$y\text{'}\text{'}\text{'}+y\text{'}\text{'}-2y=x{e}^x+1$

find a general solution to the given equation.

$y\text{'}\text{'}\text{'}-3y\text{'}\text{'}+3y\text{'}-y=e^x$

find a general solution to the given equation.

$y\text{'}\text{'}\text{'}+4y\text{'}\text{'}+y\text{'}-26y=e^{-3x}sin2x+x$

. find a differential operator that annihilates the given function.

$x^4-x^2+11$

find a differential operator that annihilates the given function.

$3x^2-6x+1$

find a differential operator that annihilates the given function.

$e^{-7x}$

find a differential operator that annihilates the given function.

$e^{5x}$

find a differential operator that annihilates the given function.

$e^{2x}-6e^x$

find a differential operator that annihilates the given function.

$x^2-e^x$

find a differential operator that annihilates the given function.

$x^2{e}^{-x}sin2x$

find a differential operator that annihilates the given function.

$x{e}^{3x}cos5x$

. find a differential operator that annihilates the given function.

$x{e}^{-2x}+x{e}^{-5x}sin3x$

find a differential operator that annihilates the given function

$x^2{e}^x-xsin4x+x^3$

use the annihilator method to determinethe form of a particular solution for the given equation.$u\text{'}\text{'}-5u\text{'}+6u=cos2x+1$

use the annihilator method to determinethe form of a particular solution for the given equation.

$y\text{'}\text{'}+6y\text{'}+8y=e^{3x}-sinx$

use the annihilator method to determinethe form of a particular solution for the given equation $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{3}\mathbf{x}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}$

use the annihilator method to determinethe form of a particular solution for the given equation. $\mathit{\theta }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{\theta }\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{x}}$

. use the annihilator method to determinethe form of a particular solution for the given equation.

$y\text{'}\text{'}-6y\text{'}+9y=sin2x+x$

use the annihilator method to determinethe form of a particular solution for the given equation.

$y\text{'}\text{'}+2y\text{'}+y=x^2-x+1$

use the annihilator method to determinethe form of a particular solution for the given equation.

$y\text{'}\text{'}+2y\text{'}+2y=e^{-x}cosx+x^2$

use the annihilator method to determinethe form of a particular solution for the given equation. $y\text{'}\text{'}-6y\text{'}+10y=e^{3x}-x$

use the annihilator method to determinethe form of a particular solution for the given equation. $z\text{'}\text{'}\text{'}-2z\text{'}\text{'}+z\text{'}=x-e^x$

use the annihilator method to determinethe form of a particular solution for the given equation.

$y\text{'}\text{'}\text{'}+2y\text{'}\text{'}-y\text{'}-2y=e^x-1$

Use the annihilator method to show that if$a_0\ne 0$in equation (4) and $f\left(x\right)$ has the form (17) $f\left(x\right)=b_m{x}^m+b_{m-1}{x}^{m-1}+\cdots +b_{1}x+b_0$, then $y_p\left(x\right)=B_{mr}{x}^m+B_{m-1}{x}^{m-1}+\cdots +B_{1}x+B_0$ is the form of a particular solution to equation (4).

Use the annihilator method to show that if${\mathit{a}}_{\mathbf{0}}\mathbf{=}\mathbf{0}$and ${\mathit{a}}_{\mathbf{1}}\mathbf{\ne }\mathbf{0}$ in (4) and has the form $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$given in (17), then equation (4) has a particular solution of the form ${\mathit{y}}_{\mathbf{p}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{x}\left\{B_m{x}^m+B_{m-1}{x}^{m-1}+\cdots +B_{1}x+B_0\right\}$

Use the annihilator method to show that if $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ in (4) has the form $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{B}{\mathit{e}}^{\mathbf{\alpha }\mathbf{x}}$, then equation (4) has a particular solution of the form ${\mathit{y}}_{\mathbf{p}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{s}}\mathit{B}{\mathit{e}}^{\mathbf{\alpha }\mathbf{x}}$, where $\mathit{s}$is chosen to be the smallest nonnegative integer such that ${\mathit{x}}^{\mathbf{s}}{\mathit{e}}^{\mathbf{\alpha }\mathbf{x}}$ is not a solution to the corresponding homogeneous equation

Use the annihilator method to show that if$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$in (4) has the form $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{a}\mathit{c}\mathit{o}\mathit{s}\mathit{\beta }\mathit{x}\mathbf{+}\mathit{b}\mathit{s}\mathit{i}\mathit{n}\mathit{\beta }\mathit{x}\mathbf{,}$

then equation (4) has a particular solution of the form

(18)${\mathit{y}}_{\mathbf{p}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{s}}\mathbf{\{}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathit{\beta }\mathit{x}\mathbf{+}\mathit{B}\mathit{s}\mathit{i}\mathit{n}\mathit{\beta }\mathit{x}\mathbf{\}}$ ,where s is chosen to be the smallest nonnegative integer such that ${\mathit{x}}^{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\mathit{\beta }\mathit{x}$ and ${\mathit{x}}^{\mathbf{3}}\mathbf{sin}\mathit{\beta }\mathit{x}$are not solutions to the corresponding homogeneous equation

In Problems 38 and 39, use the elimination method of Section to find a general solution to the given system.

$\mathit{x}\mathbf{-}{\mathit{d}}^{\mathbf{2}}\mathit{y}\mathbf{/}\mathit{d}{\mathit{t}}^{\mathbf{2}}\mathbf{=}\mathit{t}\mathbf{+}\mathbf{1}$

$\mathit{d}\mathit{x}\mathbf{/}\mathit{d}\mathit{t}\mathbf{+}\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{t}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{t}}$