Chapter 8: Ordinary Differential Equations

(a) Show that $\mathbf{D}\left(e^{\mathrm{ax}}y\right)\mathbf{=}{\mathbf{e}}^{\mathbf{ax}}\left(D+a\right)\mathbf{y}\mathbf{,}{\mathbf{D}}^{\mathbf{2}}\left(e^{\mathrm{ax}}y\right)\mathbf{=}{\mathbf{e}}^{\mathbf{ax}}{\left(D+a\right)}^{\mathbf{2}}\mathbf{y}$, and so on; that is, for any positive integral $\mathbf{n}$, ${\mathbf{D}}^{\mathbf{n}}\left(e^{\mathrm{ax}}y\right)\mathbf{=}{\mathbf{e}}^{\mathbf{ax}}{\left(D+a\right)}^{\mathbf{n}}\mathbf{y}\mathbf{.}$

Thus, show that ifis any polynomial in the operator $\mathbf{D}$, then $\mathbf{L}\left(D\right)\left(e^{\mathrm{ax}}y\right)\mathbf{=}{\mathbf{e}}^{\mathbf{ax}}\mathbf{L}\left(D+a\right)\mathbf{y}$.

This is called the exponential shift.

(b) Use to show that ${\left(D-1\right)}^{\mathbf{3}}\left(e^{x}y\right)\mathbf{=}{\mathbf{e}}^{\mathbf{x}}{\mathbf{D}}^{\mathbf{3}}\mathbf{y}\mathbf{,}\left(D^2+D-6\right)\left(e^{-3x}y\right)\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}}\left(D^2-5D\right)\mathbf{y}\mathbf{.}$.

(c) Replace $\mathbf{D}$by $\mathbf{D}\mathbf{-}\mathbf{a}$, to obtain ${\mathbf{e}}^{\mathbf{ax}}\mathbf{P}\left(D\right)\mathbf{y}\mathbf{=}\mathbf{P}\left(D-a\right){\mathbf{e}}^{\mathbf{ax}}\mathbf{y}$

This is called the inverse exponential shift.

(d) Using (c), we can change a differential equation whose right-hand side is an exponential times a

polynomial, to one whose right-hand side is just a polynomial. For example, consider

$\left(D^2-D-6\right)\mathbf{y}\mathbf{=}\mathbf{10}\mathbf{\times }{\mathbf{e}}^{\mathbf{2}\mathbf{x}}$; multiplying both sides by ${\mathbf{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}}$and using (c), we get

${\mathbf{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}}\left(D^2-D-6\right)\mathbf{y}\mathbf{=}{\left[{\left(D+3\right)}^2-\left(D+3\right)-6^{"}\right]}^{\mathbf{"}}{\mathbf{ye}}^{\mathbf{-}\mathbf{3}\mathbf{x}}\mathbf{=}\left(D^2+5D\right){\mathbf{ye}}^{\mathbf{-}\mathbf{3}\mathbf{x}}\mathbf{=}\mathbf{10}\mathbf{x}$

Show that a solution of $\left(D^2+5D\right)\mathbf{u}\mathbf{=}\mathbf{10}\mathbf{x}$is $\mathbf{u}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{5}}\mathbf{x}$; then or use this method to solve Problems 23 to 26.

Find the orthogonal trajectories of each of the following families of curves. In each case, sketch or computer plot several of the given curves and several of their orthogonal trajectories. Be careful to eliminate the constant from $\mathit{y}\mathbf{\text{'}}$for the original curves; this constant takes different values for different curves of the original family, and you want an expression for $\mathit{y}\mathbf{\text{'}}$which is valid for all curves of the family crossed by the orthogonal trajectory you are trying to find. See equations $\left(2.10\right)$to $\left(2.12\right)$

${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathbf{.}$

Solve the following sets of equations by the Laplace transform method

.$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{z}\mathbf{\text{'}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}{\mathbf{\text{'}}}_{\mathbf{0}}\mathbf{=}\mathbf{1} \\ \mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{z}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}{\mathit{e}}^{\mathbf{t}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{z}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{z}{\mathbf{\text{'}}}_{\mathbf{0}}\mathbf{=}\mathbf{1} \end{gathered}$$

Let Dstand for $\mathit{d}\mathbf{/}\mathit{d}\mathit{x}$, that is, $\mathit{D}\mathit{y}\mathbf{=}\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{x}$; then

${\mathit{D}}^{\mathbf{2}}\mathit{y}\mathbf{=}\mathit{D}\mathbf{\left(}\mathit{D}\mathit{y}\mathbf{\right)}\mathbf{=}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{,}{\mathit{D}}^{\mathbf{3}}\mathit{y}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{3}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{3}}}\mathbf{,}\mathbf{\text{etc.}}$

D(or an expression involving D) is called a differential operator. Two operators are equal if they give the same results when they operate on yFor example,

$\mathit{D}\mathbf{(}\mathit{D}\mathbf{+}\mathit{x}\mathbf{)}\mathit{y}\mathbf{=}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathbf{x}\mathbf{y}\mathbf{)}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\mathit{x}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{(}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{D}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{y}$

So, we say that

$\mathit{D}\mathbf{(}\mathit{D}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{=}{\mathit{D}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathit{D}\mathbf{+}\mathbf{1}$

In a similar way show that:

(a) $\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathbf{=}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{=}{\mathit{D}}^{\mathbf{2}}\mathbf{-}\mathbf{(}\mathit{b}\mathbf{+}\mathit{a}\mathbf{)}\mathit{D}\mathbf{+}\mathit{a}\mathit{b}$For constantand.

(b).${\mathit{D}}^{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{(}\mathit{D}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}{\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{D}\mathbf{+}\mathbf{1}\mathbf{)}$

(c)$\mathit{D}\mathit{x}\mathbf{=}\mathit{x}\mathit{D}\mathbf{+}\mathbf{1}$. (Note thatDand xdo not commute, that is,$\mathit{D}\mathit{x}\mathbf{\ne }\mathit{x}\mathit{D}$.)

(d),$\mathbf{(}\mathit{D}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{=}{\mathit{D}}^{\mathbf{2}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$but.$\mathbf{(}\mathit{D}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{=}{\mathit{D}}^{\mathbf{2}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$

Comment: The operator equations in (c) and (d) are useful in quantum mechanics; see Chapter 12, Section 22.

$\mathbf{(}\mathit{D}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{=}{\mathit{D}}^{\mathbf{2}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$

In Example 3, we used the second solution in (5.24), and obtained (5.25) as the particular solution satisfying the given initial conditions. Show that the first and third solutions in (5.24) also give the particular solution (5.25) satisfying the given initial conditions.

Solve the following sets of equations by the Laplace transform method

$$\begin{gathered} \mathit{z}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0} \\ {\mathit{y}}_{\mathbf{0}}\mathbf{=}{\mathit{z}}_{\mathbf{0}}\mathbf{=}\mathbf{0} \\ \mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{z}\mathbf{=}\mathbf{2} \end{gathered}$$

Using Problems 29 and 31b show that equation (6.24) is correct.

Find the orthogonal trajectories of each of the following families of curves. In each case, sketch or computer plot several of the given curves and several of their orthogonal trajectories. Be careful to eliminate the constant from $\mathit{y}\mathbf{\text{'}}$for the original curves; this constant takes different values for different curves of the original family, and you want an expression for $\mathit{y}\mathbf{\text{'}}$which is valid for all curves of the family crossed by the orthogonal trajectory you are trying to find. See equations $\left(2.10\right)$to $\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{12}\mathbf{)}$

$\mathit{y}\mathbf{=}\mathit{k}{\mathit{x}}^{\mathbf{2}}$

Find the orthogonal trajectories of each of the following families of curves. In each case, sketch or computer plot several of the given curves and several of their orthogonal trajectories. Be careful to eliminate the constant from $\mathit{y}\mathbf{\text{'}}$for the original curves; this constant takes different values for different curves of the original family, and you want an expression for $\mathit{y}\mathbf{\text{'}}$which is valid for all curves of the family crossed by the orthogonal trajectory you are trying to find. See equations $\left(2.10\right)$to $\left(2.10\right)$

$\mathit{y}\mathbf{=}\mathit{k}{\mathit{x}}^{\mathbf{n}}$. (Assume that n is a given number; the different curves of the family have different values of k.)

Aparticle moves along the$x$axis subject to a force toward the origin proportional to$x($say$-kx)$. Show that the particle executes simple harmonic motion (Example 3). Find the kinetic energy$\frac{1}{2}m{v}^2$and the potential energy$\frac{1}{2}k{x}^2$as functions ofand show that the total energy is constant. Find the time averages of the potential energy and the kinetic energy and show that these averages are each equal to one-half the total energy (see average values, Chapter 7, Section 4).