Differential Equations

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Differential Equations.

In this section we shall show how to obtain the general solution of any differential equation of the form

[[math]] \begin{equation} \frac{d^{2}y}{dx^2} + a \frac{dy}{dx} + by = 0, \label{eq6.8.1} \end{equation} [[/math]]

where [math]a[/math] and [math]b[/math] are real constants. Differential equations of this type occur frequently in mechanics and also in the theory of electric circuits. Equation (1) is a second- order differential equation, since it contains the second derivative [math]\frac{d^{2}y}{dx^2}[/math] but no higher derivative. It is called linear because each one of [math]y, \frac{dy}{dx}[/math], and [math]\frac{d^{2}y}{dx^2}[/math] occurs, if at all, to the first power. That is, if we set [math]\frac{dy}{dx} = z[/math] and [math]\frac{d^{2}y}{dx^2} = w[/math], then (1) becomes [math]w + az + by = 0[/math], and the left side is a linear polynomial, or polynomial of first degree, in [math]w, z[/math], and [math]y[/math]. A secondorder linear differential equation more general than (1) is

[[math]] \frac{d^{2}y}{dx^2} + f(x) \frac{dy}{dx} + g(x)y = h(x), [[/math]]

where [math]f[/math], [math]g[/math], and [math]h[/math] are given functions of [math]x[/math]. Equation (1) is a special case, called homogeneous, because [math]h[/math] is the zero function, and said to have constant coefilcients, since [math]f[/math] and [math]g[/math] are constant functions. Thus the topic of this section becomes: the study of second-order, linear, homogeneous differential equations with constant coefficients. An important and easily proved property of differential equations of this kind is the following:

Theorem

If [math]y_1[/math] and [math]y_2[/math] are any two solutions of the differential equation (1), and if [math]c_1[/math] and [math]c_2[/math] are any two real numbers, then [math]c_{1}y_{1} + c_{2}y_{2}[/math] is also a solution.

Show Proof

The proof uses only the elementary properties of the derivative. We know that

[[math]] \frac{d}{dx}(c_{1}y_{1} + c_{2}y_{2}) = c_{1} \frac{dy_1}{dx} + c_{2} \frac{dy_2}{dx}. [[/math]]

Hence

[[math]] \begin{eqnarray*} \frac{d^2}{{dx}^2} (c_{1}y_{1} + c_{2}y_{2}) &=& \frac{d}{dx}\Bigl(c_{1} \frac{dy_1}{dx} + c_{2} \frac{dy_2}{dx}\Bigr) \\ &=& c_{1} \frac{d^{2}y_1}{{dx}^2} + c_{2} \frac{d^{2}y_2}{{dx}^2}. \end{eqnarray*} [[/math]]

To test whether or not [math]c_{1}y_{1} + c_{2}y_{2}[/math] is a solution, we substitute it for [math]y[/math] in the differential equation:

[[math]] \begin{eqnarray*} && \frac{d^{2}}{dx^2} (c_{1}y_{1} + c_{2}y_{2}) + a \frac{d}{dx} (c_{1}y_{1} + c_{2}y_{2}) + b(c_{1}y_{1} + c_{2}y_{2})\\ &=& c_{1} \frac{d^{2}y_1}{dx^2} + c_{2} \frac{d^{2}y_2}{dx^2} + ac_{1} \frac{dy_1}{dx} + ac_{2} \frac{dy_2}{ dx} + bc_{1}y_{1} + bc_{2}y_{2}\\ &=& c_{1}\Bigl(\frac{d^{2}y_1}{dx^2} + a \frac{dy_1}{dx} + by_1\Bigr) + c_{2}\Bigl(\frac{d^{2}y_2}{dx^2} + a \frac{dy_2}{dx} + by_2\Bigr). \end{eqnarray*} [[/math]]

The expressions in parentheses in the last line are both zero because, [math]y_1[/math] and [math]y_2[/math] are by assumption solutions of the differential equation. Hence the top line is also zero, and so [math]c_{1}y_{1} + c_{2}y_{2}[/math] is a solution. This completes the proof.

■

It follows in particular that the sum and difference of any two solutions of (1) is a solution, and also that any constant multiple of a solution is again a solution. Finally, note that the constant function 0 is a solution of (1) for any constants [math]a[/math] and [math]b[/math]. In Section 5 of Chapter 5 we found that the general solution of the differential equation [math]\frac{dy}{dx} + ky = 0[/math] is the function [math]y = ce^{-kx}[/math], where [math]c[/math] is an arbitrary real number. This differential equation is first-order, linear, homogeneous, and with constant coefficients. Let us see whether by any chance an exponential function might also be a solution of the second-order differential equation [math]\frac{d^{2}y}{dx^2} + a \frac{dy}{dx} + by = 0[/math]. Let [math]y = e^{rx}[/math], where [math]r[/math] is any real number. Then

[[math]] \frac{dy}{dx} = re^{rx}, \;\;\; \frac{d^{2}y}{dx^2} = r^{2} e^{rx} . [[/math]]

Hence

[[math]] \begin{eqnarray*} \frac{d^{2}y}{dx^2} + a \frac{dy}{dx} + by &=& r^{2}e^{rx} + are^{rx} + be^{rx} \\ &=& (r^2 + ar + b)er^{rx}. \end{eqnarray*} [[/math]]

Since [math]e^{rx}[/math] is never zero, the right side is zero if and only if [math]r^2 + ar + b = 0[/math]. That is, we have shown that

Theorem

The function [math]e^{rx}[/math] is a solution of [math]\frac{d^{2}y}{dx^2} + a \frac{dy}{dx} + by = 0[/math] if and only if the real number [math]r[/math] is a solution of [math]t^2 + at + b = 0[/math].

The latter equation is called the characteristic equation of the differential equation. \medskip Example

Consider the differential equation [math]\frac{d^{2}y}{dx^2} - \frac{dy}{dx} - 6y = 0[/math]. Its characteristic equation is [math]t^2 - t - 6 = 0[/math]. Since [math]t^2 - t - 6 = (t - 3)(t + 2)[/math], the two solutions, or roots, are 3 and -2. Hence, by (8.2), both functions [math]e^{3x}[/math] and [math]e^{-2x}[/math] are solutions of the differential equation. It follows by (8.1) that the function

[[math]] y = c_{1} e^{3x} = + c_{2}e^{-2x} [[/math]]

is a solution for any two real numbers [math]c_1[/math] and [math]c_2[/math].

The form of the general solution of the differential equation

[[math]] \frac{d^{2}y}{dx^{2}} + a \frac{dy}{dx} + dx + by = 0 [[/math]]

depends on the roots of the characteristic equation [math]t^2 + at + b = 0[/math]. There are three different cases to be considered. \medskip Cuse 1. The characteristic equation has distinct real roots. This is the simplest case. We have

[[math]] t^2 + at + b = (t - r_1)(t - r_2), [[/math]]

where [math]r_1[/math], and [math]r_2[/math] are real numbers and [math]r_{1} \neq r_2[/math]. Both functions [math]e^{r_{1}x}[/math] and [math]e^{r_{2}x}[/math] are solutions of the differential equation, and so is any linear combination [math]c_{1}e^{r_{1}x} + c_{2}e^{r_{2}x}[/math]. Moreover it can be shown, although we defer the proof until Chapter 11, that if [math]y[/math] is any solution of the differential equation, then

[[math]] \begin{equation} y = c_{1}e^{r_{1}x} + c_{2}e^{r_{2}x} , \label{eq6.8.2} \end{equation} [[/math]]

for some two real numbers [math]c_1[/math] and [math]c_2[/math]. Hence we say that (2) is the general solution. In Example 1 the function [math]c_{1}e^{3x} + c_{2}e^{-2x}[/math] is therefore the general solution of the differential equation [math]\frac{d^{2}y}{dx^2} - \frac{dy}{dx} - 6y = 0[/math]. \medskip Case 2. The characteristic equation has complex roots. The roots of [math]t^2 + at + b = 0[/math] are given by the quadratic formula

[[math]] r_{1}, r_{2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}. [[/math]]

Since [math]a[/math] and [math]b[/math] are real, [math]r_1[/math] and [math]r_2[/math] are complex if and only if [math]a^2 - 4b \lt 0[/math], which we now assume. Setting [math]\alpha = - \frac{a}{2}[/math] and [math]\beta = \frac{\sqrt{4b - a^2}}{2}[/math], we have

[[math]] r_1 = \alpha + i\beta, \;\;\; r_2 = \alpha - i\beta. [[/math]]

Note that [math]r_1[/math], and [math]r_2[/math] are complex conjugates of each other. Motivated by the situation in Case 1, in which [math]r_1[/math] and [math]r_2[/math] were real, we consider the complex-valued function [math]c_{1}e^{r_{1}x} + c_{2}e^{r_{2}x}[/math], where we now allow [math]c_1[/math] and [math]c_2[/math], to be complex numbers. We shall show that

Theorem

If [math]c_1[/math] and [math]c_2[/math] are any two complex conjugates of each other and if [math]r_1[/math] and [math]r_2[/math] are complex solutions of the characteristic equation, then the function

[[math]] y = c_{1}e^{r_{1}x} + c_{2}e^{r_{2}x} [[/math]]

is real-valued. Moreover it is a solution of the differential equation (1).

Show Proof

Let [math]c_1 = \gamma + i\delta[/math] and [math]c_2 = \gamma - i\delta[/math]. Since [math]r_1 = \alpha + i\beta[/math] and [math]r_2 = \alpha - i\beta[/math], we have

[[math]] \begin{eqnarray*} c_{1}e^{r_{1}x} + c_{2}e^{r_{2}x} &=& (\gamma + i\delta) e^{(\alpha + i\beta)x} + (\gamma - i \delta) e^{(\alpha - i\beta)x}\\ &=& e^{\alpha x}[(\gamma + i \delta) e^{i\beta x} + (\gamma - i\delta) e^{-i\beta x}]. \end{eqnarray*} [[/math]]

Recall that [math]e^{i\beta x} = \cos \beta x + i \sin \beta x[/math] and [math]e^{-i\beta x} = \cos(\beta x) + i sin(-\beta x) = \cos \beta x - i \sin \beta x[/math]. Substituting, we get

[[math]] \begin{eqnarray*} c_{1}e^{r_{1}x} + c_{2}e^{r_{2}x} &=& e^{\alpha x} [(\gamma + i\delta)(\cos \beta x + i \sin \beta x) + (\gamma - i\delta)(\cos \beta x - i \sin \beta x)]\\ &=& e^{\alpha x} (2\gamma \cos \beta x - 2\delta \sin \beta x). \end{eqnarray*} [[/math]]

The right side is certainly real-valued, and this proves the first statement of the theorem. Since [math]\gamma[/math] and [math]\delta[/math] are arbitrary real numbers, so are [math]2\gamma[/math] and [math]-2\delta[/math]. We may therefore replace [math]2\gamma[/math] by [math]k_1[/math] and [math]-2\delta[/math] by [math]k_2[/math]. We prove the second statement of the theorem by showing that the function

[[math]] \begin{equation} y = e^{\alpha x}(k_{1} \cos \beta x + k_{2} \sin \beta x) \label{eq6.8.3} \end{equation} [[/math]]

is a solution of the differential equation. Let [math]y_1 = e^{\alpha x} \cos \beta x[/math] and [math]y_2 = e^{ax} \sin \beta x[/math]. Since [math]y = k_{1}y_1 + k_{2}y_2[/math], it follows by (8.1) that it is enough to show that [math]y_1[/math] and [math]y_2[/math] separately are solutions of the differential equation. We give the proof for [math]y_1[/math] and leave it to the reader to check it for [math]y_2[/math]. By the product rule,

[[math]] \begin{eqnarray*} \frac{dy_1}{dx} &=& \frac{d}{dx} (e^{\alpha x}\cos \beta x) = \alpha e^{\alpha x} \cos \beta x - \beta e^{\alpha x} \sin \beta x \\ &=& e^{\alpha x}(\alpha \cos \beta x - \beta \sin \beta x). \end{eqnarray*} [[/math]]

Hence

[[math]] \begin{eqnarray*} \frac{d^{2}y_1}{dx^2} &=& \alpha e^{\alpha x}(\alpha \cos \beta x - \beta \sin \beta x) + e^{\alpha x}(-\alpha \beta \sin \beta x - \beta^{2} \cos \beta x) \\ &=& e^{\alpha x} [(\alpha^2 - \beta^2) \cos \beta x - 2\alpha\beta \sin \beta x]. \end{eqnarray*} [[/math]]

Thus

[[math]] \begin{eqnarray*} \frac{d^{2}y_1}{dx^2} + a \frac{dy_1}{dx} + by_1 &=& e^{\alpha x}[(\alpha^2 - \beta^2) \cos \beta x - 2\alpha \beta \sin \beta x] \\ & & + ae^{\alpha x}(\alpha \cos \beta x - \beta \sin \beta x) + be^{\alpha x} \cos \beta x \\ &=& e^{\alpha x}([(\alpha^2 - \beta^2) + a\alpha + b] \cos \beta x - \beta (2\alpha + a) \sin \beta x). \end{eqnarray*} [[/math]]

But, remembering that [math]r_1 = \alpha + i \beta[/math] and [math]r_2 = \alpha - i \beta[/math] and that these are the roots of the characteristic equation, we read from the quadratic formula that

[[math]] \alpha = - \frac{a}{2}, \;\;\; \beta = \frac{\sqrt{4b - a^2}}{2}. [[/math]]

Hence

[[math]] \begin{eqnarray*} (\alpha^2 - \beta^2) + a \alpha + b &=& \Bigl( -\frac{a}{2} \Bigr)^2 - \Bigl( \frac{\sqrt{4b - a^2}}{2} \Bigr)^2 + a \Bigl(- \frac{a}{2} \Bigr) + b\\ &=& \frac{a^2}{4} - b + \frac{a^2}{4} -\frac{a^2}{2} + b = 0, \end{eqnarray*} [[/math]]

and also

[[math]] 2 \alpha + a = 2 \Bigl(-\frac{a}{2} \Bigr) + a = -a + a = 0, [[/math]]

whence we get

[[math]] \begin{eqnarray*} \frac{d^{2}y_1}{dx^2} + \alpha \frac {dy_1}{dx} + by_1 &=& e^{\alpha x} (0 \cdot \cos \beta x - \beta \cdot 0 \cdot \sin \beta x) \\ &=& 0, \end{eqnarray*} [[/math]]

and so [math]y_1[/math] is a solution. Assuming the analogous proof for [math]y_2[/math], it follows that [math]y[/math], as defined by (3), is also a solution and the proof is complete.

■

It can be shown, although again we defer the proof, that if [math]y[/math] is any real solution to the differential equation (1), and if the roots [math]r_1[/math] and [math]r_2[/math] of the characteristic equation are complex, then

[[math]] \begin{equation} y = c_{1}e^{r_{1}x} + c_{2}e^{r_{2}x} , \label{eq6.8.4} \end{equation} [[/math]]

for some complex number [math]c_1[/math] and its complex conjugate [math]c_2[/math]. Hence, if the roots are complex, the general solution of the differential equation can be written either as (4), or in the equivalent form,

[[math]] \begin{equation} y = e^{\alpha x}(k_{1} \cos \beta x + k_{2} \sin \beta x), \label{eq6.8.5} \end{equation} [[/math]]

where [math]r_1 = \alpha + i\beta[/math] and [math]r_2 = a - i \beta[/math], and [math]k_1[/math] and [math]k_2[/math] are arbitrary real numbers. Note that solutions (2) and (4) look the same, even though they involve different kinds of [math]r[/math]'s and different kinds of [math]c[/math]'s.

Example

Find the general solution of the differential equation

[[math]] \frac{d^{2}y}{dx^2} + 4 \frac{dy}{dx} + 13y = 0. [[/math]]

The characteristic equation is [math]t^2 + 4t + 13 = 0[/math]. Using the quadratic formula, we find the roots

[[math]] \begin{eqnarray*} r_1, r_2 &=& \frac{- 4 \pm \sqrt{16 - 4 \cdot 13}}{2} = \frac{- 4 \pm \sqrt{-36}}{2}\\ &=& -2 \pm 3i. \end{eqnarray*} [[/math]]

Hence, by (4), the general solution can be written

[[math]] y = c_{1}e^{(-2+3i)x}= + c_{2}e^{(-2-3i)x}, [[/math]]

where [math]c_1[/math] and [math]c_2[/math] are complex conjugates of each other. Unless otherwise stated, however, the solution should appear as an obviously real-valued function. That is, it should be written without the use of complex numbers as in (5). Hence the preferred form of the general solution is

[[math]] y = e^{-2x}(k_{1} \cos 3x + k_{2} \sin 3x). [[/math]]

We now consider the remaining possibility. \medskip Case 3. The characteristic equation [math]t^2 + at + b = 0[/math] has only one root [math]r[/math]. In this case, we have [math]t^2 + at + b = (t - r)(t - r)[/math], and the quadratic formula yields [math]r = - \frac{a}{2}[/math] and [math]\sqrt{a^2 - 4b} = 0[/math]. Theorem (8.2) is still valid, of course, and so one solution of the differtial equation [math]\frac{d^{2}y}{dx^2} + a \frac{dy}{dx} + by = 0[/math] is obtained by taking [math]y = e^{rx}[/math]. We shall show that, in the case of only one root, [math]xe^{rx}[/math] is also a solution. Setting [math]y = xe^{rx}[/math], we obtain

[[math]] \begin{eqnarray*} \frac{dy}{dx} &=& e^{rx} + xre^{rx} = e^{rx} (1 + rx),\\ \frac{d^{2}y}{dx^2} &=& re^{rx}(1 + rx) + e^{rx} \cdot r\\ &=& re^{rx} (2 + rx). \end{eqnarray*} [[/math]]

Hence

[[math]] \begin{eqnarray*} \frac{d^{2}y}{dx^2} + a \frac{dy}{dx} + by &=& re^{rx}(2 + rx) + ae^{rx}(1 + rx) + bxe^{rx}\\ &=& e^{rx}(2r + r^{2}x + a + arx + bx) \\ &=& e^{rx}[x(r^2 + ar + b) + (a + 2r)]. \end{eqnarray*} [[/math]]

Since [math]r[/math] is a root of [math]t^2 + at + b[/math], we know that [math]r^2 + ar + b = 0[/math]. Moreover, we have seen that [math]r = -\frac{a}{2}[/math], and so [math]a + 2r = 0[/math]. It follows that the last expression in the above equations is equal to zero, which shows that the function [math]xe^{rx}[/math] is a solution of the differential equation. Thus [math]e^{rx}[/math] is one solution, and [math]xe^{rx}[/math] is another. It follows by (8.1) that, for any two real numbers [math]c_1[/math] and [math]c_2[/math], a solution is given by

[[math]] y = c_{1}xe^{rx} + c_{2}e^{rx} = (c_{1}x + c_{2})e^{rx}, [[/math]]

Conversely, it can be shown that if [math]y[/math] is any solution of the differential equation (1), and if the characteristic equation has only one root [math]r[/math], then

[[math]] \begin{equation} y = (c_{1}x + c_2)e^{rx} \label{eq6.8.6} \end{equation} [[/math]]

for some pair of real numbers [math]c_1[/math] and [math]c_2[/math]. The general solution in the case of a single root is therefore given by (6). Example Find the general solution of the differential equation [math]9y'' - 6y' + y = 0[/math]. Here we have used the common notation [math]y'[/math] and [math]y''[/math] for the first and second derivatives of the unknown function [math]y[/math]. Dividing the equation by 9 to obtain a leading coefficient of 1, we get [math]y'' - \frac{2}{3}y' + \frac{1}{9}y = 0[/math], for which the characteristic equation is [math]t^2 - \frac{2}{3}t + \frac{1}{9} = 0[/math]. Since [math]t^2 - \frac{2}{3}t + \frac{1}{9} = (t - \frac{1}{3})(t - \frac{1}{3})[/math], there is only one root, [math]r = 3[/math]. Hence

[[math]] y = (c_{1}x + c_{2})e^{x/3} [[/math]]

is the general solution.

The solution of a differential equation can be checked just as simply as an indefinite integral, by differentiation and substitution.

\end{exercise}

General references

Doyle, Peter G. (2008). "Crowell and Slesnick's Calculus with Analytic Geometry" (PDF). Retrieved Oct 29, 2024.