Linear Homogeneous Differential Equations

The full description of these equations is: Linear constant coefficient homogeneous equations. The equations described in the title have the form

$$a_n y^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = 0.$$

Here y is a function of x, and $a_n$ , ... , $a_0$ are {\it constants}. Linear means the equation is a sum of the derivatives of y, each multiplied by x stuff. (In this case, the x stuff is constant.) Homogeneous means that the right side is 0 --- there's no term involving only x.

It's convenient to let $D
   = \der {} x$ stand for the operation of differentiating with respect to x. (Note that $D = \der {} x$ is the operation of differentiation, whereas $Dy = \der y x$ is the derivative.) In this notation, $D^2$ computes the second derivative, $D^3$ computes the third derivative, and so on. The equation above becomes

$$(a_nD^n + \cdots a_2D^2 + a_1D + a_0)y = 0.$$


Example. The following equations are linear homogeneous equations with constant coefficients:

$$y''' + 3y'' + 3y' + y = 0,$$

$$y'' - 5y' - 6y = 0,$$

$$((D - 1)(D - 2)(D - \pi)y = 0.\quad\halmos$$


A solution to the equation is a function $y = f(x)$ which satisfies the equation. Equivalently, if you think of $a_nD^n
   + \cdots a_2D^2 + a_1D + a_0$ as a linear transformation, it is an element of the kernel of the transformation.

The general solution is a linear combination of the elements of a basis for the kernel, with the coefficients being arbitrary constants.

The form of the equation makes it reasonable that a solution should be a function whose derivatives are constant multiples of itself. $e^{mx}$ is such a function:

$$\der {} x e^{mx} = re^{mx}, \quad \dfrac{d}{dx^2} e^{mx} = m^2 e^{mx}, \quad \ldots, \quad \dfrac{d^n}{dx^n} e^{mx} =m^n e^{mx}.$$

Plug $e^{mx}$ into

$$y^{(n)} + b_{n-1} y^{(n-1)} + \cdots + b_2 y'' + b_1 y' + b_0 y = 0.$$

The result:

$$m^n e^{mx} + b_{n-1} m^{n-1} e^{mx} + \cdots + b_2 m^2 e^{mx} + b_1 m e^{mx} + b_0 e^{mx} = 0.$$

Factor out $e^{mx}$ and cancel it. This leaves

$$m^n + b_{n-1} m^{n-1} + \cdots + b_2 m^2 + b_1 m + b_0 = 0.$$

Thus, $e^{mx}$ is a solution to the original equation exactly when m is a root of this polynomial. The polynomial is called the characteristic polynomial; as the derivation showed, it's obtained by building a polynomial using the coefficients of the original differential equation.


Example. Solve $y'' - 5y' - 6y = 0$ .

The characteristic polynomial is $m^2 - 5m - 6$ ; solving $m^2 - 5m - 6 = 0$ yields $(m - 6)(m + 1) = 0$ , so $m = 6$ or $m =
   -1$ . The general solution is

$$y = c_1 e^{6x} + c_2 e^{-x}.$$

You can check this by plugging back in. Here are the derivatives:

$$y = c_1 e^{6x} + c_2 e^{-x}, \quad y' = 6c_1 e^{6x} - c_2 e^{-x}, \quad y'' = 36c_1 e^{6x} + c_2 e^{-x}.$$

Therefore,

$$y'' - 5y' + 6y = \left(36c_1 e^{6x} + c_2 e^{-x}\right) - 5\left(6c_1 e^{6x} - c_2 e^{-x}\right) - 6\left(c_1 e^{6x} + c_2 e^{-x}\right) = 0.\quad\halmos$$


Example. Solve $(D^4 - 9D^2 + 20)y = 0$ .

It's easy to write down the characteristic equation: just replace the D's with m's:

$$m^4 - 9m^2 + 20 = 0, \quad (x^2 - 4)(x^2 - 5) = 0, \quad (x - 2)(x + 2)(x - \sqrt{5})(x + \sqrt{5}) = 0.$$

The roots are $\pm 2$ and $\pm \sqrt{5}$ . (Don't fall into the trap of assuming that roots must be integers, or even rationals!) The solution is

$$y = c_1 e^{2x} + c_2 e^{-2x} + c_3 e^{\sqrt{5}x} + c_4 e^{-\sqrt{5}x}.\quad\halmos$$


What happens if there are repeated roots? Look at the equation $y'' - 2y' + y = 0$ . The characteristic equation is $x^2 - 2x + 1 = 0$ , which has $x = 1$ as a double root. It is true that $e^x$ is a solution, but it would be incorrect to write $y = c_1 e^x + c_2 e^x$ .

The terms $c_1 e^x$ and $c_2 e^x$ are redundant --- you could combine them to get $y = (c_1 + c_2)e^x = c_3
   e^x$ . To put it another way, as function $e^x$ and $e^x$ are linearly dependent.

It is reasonable to suppose that for a second order equation you should have two different solutions. $e^x$ is one; how can you find another?

The idea is to guess the form that such a solution might take. Guess:

$$y = f(x) e^x,$$

i.e. something times the known solution $e^x$ . What should f be?

To find f, plug $f(x)
   e^x$ into the equation. Here are the derivatives:

$$y = f(x) e^x, \quad y' = f'(x) e^x + f(x) e^x, \quad y'' = f''(x) e^x + 2f'(x) e^x + f(x) e^x.$$

Plug them in:

$$y'' - 2y' + y = (f''(x) e^x + 2f'(x) e^x + f(x) e^x) - 2(f'(x) e^x + f(x) e^x) + f(x) e^x = f''(x) e^x = 0.$$

Hence, $f''(x) = 0$ . Integrate twice and obtain $f(x) = c_1 + c_2 x$ . Thus,

$$y = c_1 e^x + c_2x e^x.$$

In fact, this is the general solution --- notice the two arbitrary constants. The functions $e^x$ and $x e^x$ are indpendent solutions to the original equation.

In general, if m is a repeated root of multiplicity k in the characteristic polynomial, you get terms $e^{mx}$ , $xe^{mx}$ , ... , $x^{k-1} e^{mx}$ in the general solution.


Example. Solve $\dfrac{d^3y}{dx^3} - 3
   \dfrac{d^2y}{dx^2} + 3 \der y x - y = 0$ .

The characteristic equation is $m^3 - 3m^2 + 3m - 1 = 0$ , which has $m = 1$ as a root with multiplicity 3. The general solution is

$$y = c_1 e^x + c_2 x e^x + c_3 x^2 e^x.\quad\halmos$$


Example. Solve $(D^2 - 1)(D^2 - D - 2)y =
   0$ .

The characteristic equation is $(m^2 - 1)(m^2 - m - 2) = 0$ , or $(m - 1)(m + 1)^2(m - 2) =
   0$ . The roots are 1, 2, and -1 (double). The general solution is

$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{-x} + c_4 x e^{-x}.$$

Note: You can write the terms in the solution in any order you please. Nor does it matter which "c" goes with which term, since they {\it are} arbitrary constants.


Example. ( Linear systems) Suppose x and y are functions of t. Consider the system of differential equations

$$\der x t = x' = x + 4y, \quad \der y t = y' = 2x + 3y.$$

I want to solve for x and y in terms of t.

Solve the second equation for x:

$$x = \dfrac{1}{2}\left(y' - 3y\right).$$

Differentiate:

$$x' = \dfrac{1}{2}\left(y'' - 3y'\right).$$

Plug the expressions for x and $x'$ into the first equation:

$$\dfrac{1}{2}\left(y'' - 3y'\right) = \dfrac{1}{2}\left(y' - 3y\right) + 4y.$$

Simplify:

$$y'' - 4y' - 5y = 0.$$

The characteristic equation is $m^2 - 4m - 5 = 0$ , or $(m - 5)(m + 1) = 0$ . The roots are $m = 5$ and $m =
   -1$ . Therefore,

$$y = c_1 e^{5t} + c_2 e^{-t}.$$

Now $y' = 5c_1 e^{5t} -
   c_2 e^{-t}$ , so

$$x = \dfrac{1}{2}\left(y' - 3y\right) = \dfrac{1}{2}\left[\left(5c_1 e^{5t} - c_2 e^{-t}\right) - 3\left(c_1 e^{5t} + c_2 e^{-t}\right)\right] = c_1 e^{5t} - 2c_2 e^{-t}.$$

There are other ways of solving linear systems, but for small systems brute force works reasonably well!


Now suppose the characteristic equation has a complex root $a + bi$ . From basic algebra, complex roots of real polynomials come in conjugate pairs: $a + bi$ and $a -
   bi$ . It's reasonable to expect solutions

$$c_1 e^{(a+bi)x} \quad\hbox{and}\quad c_2 e^{(a-bi)x}.$$

However, these are complex solutions, and you should have real solutions to the original real differential equation. I'll use the complex exponential formula

$$e^{i\theta} = \cos \theta + i \sin \theta, \quad \theta \in \real.$$

You can derive this formula by considering the Taylor series for $e^{i\theta}$ , $\cos \theta$ , and $\sin \theta$ .

Now

$$c_1 e^{(a+bi)x} + c_2 e^{(a-bi)x} = c_1 e^{ax} e^{ibx} + c_2 e^{ax} e^{-ibx} =$$

$$e^{ax} \left(c_1 (\cos bx + i \sin bx) + c_2 (\cos bx - i \sin bx)\right) = e^{ax} \left( (c_1 + c_2)\cos bx + i (c_1 - c_2) \sin bx\right).$$

Let $c_3 = c_1 + c_2$ and $c_4 = i(c_1 - c_2)$ . Observe that $c_1$ and $c_2$ can be solved for in terms of $c_3$ and $c_4$ , so no generality is lost with this substitution. Then

$$c_1 e^{(a+bi)x} + c_2 e^{(a-bi)x} = e^{ax} (c_3 \cos bx + c_4 \sin bx).$$

Each pair of conjugate complex roots $a \pm bi$ in the characteristic equation generates a pair of independent solutions of this form.


Example. Solve $y'' + y = 0$ .

The characteristic equation $m^2 + 1 = 0$ has roots $\pm i$ . The solution is

$$y = c_1 \cos x + c_2 \sin x.\quad\halmos$$


Example. Solve $(D^2 + 2D + 5)y = 0$ .

The characteristic equation $m^2 + 2m + 5 = 0$ has roots $m = -1 \pm 2i$ . The solution is

$$y = e^{-x} (c_1 \cos 2x + c_2 \sin 2x).\quad\halmos$$


Example. Solve $(D^3 + 1)y = 0$ .

The characteristic polynomial $m^3 + 1$ factors into $(m + 1)(m^2 - m + 1)$ . The roots are $m = -1$ and $m
   = \dfrac{1}{2} \pm i \dfrac{\sqrt{3}}{2}$ . The solution is

$$y = c_1 e^{-x} + e^{x/2}\left(c_2 \cos \dfrac{\sqrt{3}}{2}x + c_3 \sin \dfrac{\sqrt{3}}{2}x\right).\quad\halmos$$


Example. Solve $(D^2 + 4)^2y = 0$ .

The characteristic equation $(m^2 + 4)^2 = 0$ has repeated complex roots: $m = \pm 2i$ (each double). The solution is

$$y = c_1 \cos 2x + c_2 \sin 2x + c_3 x \cos 2x + c_4 x \sin 2x.$$

(What's the solution to $(D^2 + 2D + 5)^2y = 0$ ?)


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