Homogeneous linear differential equation with constant coefficients: Difference between revisions

Latest revision as of 00:03, 7 July 2012

Definition

A homogeneous linear differential equation with constant coefficients, which can also be thought of as a linear differential equation that is simultaneously an autonomous differential equation, is a differential equation of the form:

$y^{(k)} + p_{k - 1} y^{(k - 1)} + \dots + p_{1} y^{'} + p_{0} y = 0$

where $p_{0}, p_{1}, \dots, p_{k - 1}$ are all constants (i.e., real numbers). Here $y$ is the dependent variable and $x$ (which does not appear explicitly above) is the independent variable with respect to which the differentiations occur.

Solution method

Consider the following polynomial:

$t^{k} + p_{k - 1} t^{k - 1} + \dots + p_{1} t + p_{0}$

This polynomial is called the characteristic polynomial of the differential equation. We consider various cases:

Case	Solution in that case
The polynomial has pairwise distinct real roots $α_{1}, α_{2}, \dots, α_{k}$	The solution space has basis $e^{α_{1} x}, e^{α_{2} x}, \dots, e^{α_{k} x}$ . In other words, the general solution is $C_{1} e^{α_{1} x} + C_{2} e^{α_{2} x} + \dots + C_{k} e^{α_{k} x}$ where $C_{1}, C_{2}, \dots, C_{k}$ are freely varying real parameters.
The polynomial splits completely into linear factors over the reals, but with possible repetitions. $α_{1}$ occurs $s_{1}$ times, $α_{2}$ occurs $s_{2}$ times, and so on till $α_{r}$ , which occurs $s_{r}$ times. We have $s_{1} + s_{2} + \dots + s_{r} = k$ .	The solution space has basis all functions of the form $x^{d} e^{α_{i} x}$ where $0 \leq d < s_{i}$ with $d$ an integer. Thus, for each $i$ , there are $s_{i}$ basis vectors corresponding to $α_{i}$ . We get a total of $k$ basis vectors.
The polynomial splits completely over the complex numbers into distinct linear factors, but some of the roots are not real	For any real root $α$ , use $e^{α x}$ as a basis vector. Non-real roots occur in complex conjugate pairs. For a pair $a \pm i b$ , choose the vectors $e^{a x} \cos (b x)$ and $e^{a x} \sin (b x)$ . Combining, we get a basis of $k$ vectors.
The general case	For a real root $α$ of multiplicity $s$ , the $s$ basis vectors are $x^{d} e^{α x}, 0 \leq d < s$ . For a pair of complex conjugates $a \pm i b$ of multiplicity $s$ , the $2 s$ basis vectors are $x^{d} e^{a x} \cos (b x), 0 \leq d < s$ and $x^{d} e^{a x} \sin (b x), 0 \leq d < s$ .

Examples

Examples with distinct real roots

Consider the differential equation:

$y^{″} - 3 y^{'} + 2 y = 0$

The characteristic polynomial is:

$t^{2} - 3 t + 2$

We want to find the roots of this polynomial. The roots are $α_{1} = 1, α_{2} = 2$ . Thus, the general solution is:

$y = C_{1} e^{x} + C_{2} e^{2 x}, C_{1}, C_{2} \in R$

Here is another example:

$y^{‴} + 2 y^{″} - 2 y^{'} - 4 y = 0$

The characteristic polynomial is:

$t^{3} + 2 t^{2} - 2 t - 4$

The roots are $α_{1} = - 2, α_{2} = - \sqrt{2}, α_{3} = \sqrt{2}$ . The general solution is thus:

$y = C_{1} e^{- 2 x} + C_{2} e^{- \sqrt{2} x} + C_{3} e^{\sqrt{2} x}, C_{1}, C_{2}, C_{3} \in R$

Examples with repeated real roots

Consider the differential equation:

$y^{″} = 6 y^{'} - 9 y$

First, we bring everything to one side, to get:

$y^{″} - 6 y^{'} + 9 y = 0$

The characteristic polynomial is:

$t^{2} - 6 t + 9$

This factors as $(t - 3)^{2}$ , so it has a root 3 with multiplicity 2. The general solution is thus:

$y = (C_{1} + C_{2} x) e^{3 x}, C_{1}, C_{2} \in R$

Examples with real and complex roots without repetitions

Consider the differential equation:

$y^{‴} = y$

First, we bring everything to one side, to get:

$y^{‴} - y = 0$

The characteristic polynomial is:

$t^{3} - 1$

The only real root is 1, corresponding to a solution basis element $e^{x}$ . The complex roots are the roots of $t^{2} + t + 1$ , and they are complex conjugates:

$\frac{- 1 \pm \sqrt{3} i}{2}$

The basis solutions for this pair are thus $e^{- x / 2} \cos (\sqrt{3} x / 2)$ and $e^{- x / 2} \sin (\sqrt{3} x / 2)$ . The general solution is thus:

$y = C_{1} e^{x} + C_{2} e^{- x / 2} \cos (\sqrt{3} x / 2) + C_{3} e^{- x / 2} \sin (\sqrt{3} x / 2)$

@@ Line 5: / Line 5: @@
 <math>y^{(k)} + p_{k-1}y^{(k-1)} + \dots + p_1y' + p_0y = 0</math>
-where <math>p_0,p_1,\dots,p_{k-1}</math> are all constants (i.e., real numbers).
+where <math>p_0,p_1,\dots,p_{k-1}</math> are all constants (i.e., real numbers). Here <math>y</math> is the dependent variable and <math>x</math> (which does not appear explicitly above) is the independent variable with respect to which the differentiations occur.
 ===Solution method===
@@ Line 11: / Line 11: @@
 Consider the following polynomial:
-<math>t^k + p_{k-1}t^{k-1} + \dots + p_1t +p_0 = 0</math>
+<math>t^k + p_{k-1}t^{k-1} + \dots + p_1t +p_0</math>
 This polynomial is called the ''characteristic polynomial'' of the differential equation. We consider various cases:
@@ Line 24: / Line 24: @@
 | The polynomial splits completely over the complex numbers into distinct linear factors, but some of the roots are not real || For any real root <math>\alpha</math>, use <math>e^{\alpha x}</math> as a basis vector. Non-real roots occur in complex conjugate pairs. For a pair <math>a \pm ib</math>, choose the vectors <math>e^{ax}\cos(bx)</math> and <math>e^{ax}\sin(bx)</math>. Combining, we get a basis of <math>k</math> vectors.
 |-
-| The general case || {{fillin}} -- combine ideas of two preceding generalizations.
+| The general case || For a real root <math>\alpha</math> of multiplicity <math>s</math>, the <math>s</math> basis vectors are <math>x^de^{\alpha x}, 0 \le d < s</math>. For a pair of complex conjugates <math>a \pm ib</math> of multiplicity <math>s</math>, the <math>2s</math> basis vectors are <math>x^de^{ax}\cos(bx), 0 \le d < s</math> and <math>x^de^{ax}\sin(bx), 0 \le d < s</math>.
 |}
+==Examples==
+===Examples with distinct real roots===
+Consider the differential equation:
+<math>y'' - 3y' + 2y = 0</math>
+The characteristic polynomial is:
+<math>t^2 - 3t + 2</math>
+We want to find the roots of this polynomial. The roots are <math>\alpha_1 = 1, \alpha_2 = 2</math>. Thus, the general solution is:
+<math>y = C_1e^x + C_2e^{2x}, \qquad C_1,C_2 \in \R</math>
+Here is another example:
+<math>y''' + 2y'' -2y' - 4y = 0</math>
+The characteristic polynomial is:
+<math>t^3 + 2t^2 - 2t - 4</math>
+The roots are <math>\alpha_1 = -2, \alpha_2 = -\sqrt{2}, \alpha_3 = \sqrt{2}</math>. The general solution is thus:
+<math>y = C_1e^{-2x} + C_2e^{-\sqrt{2} x} + C_3e^{\sqrt{2} x}, \qquad C_1,C_2,C_3 \in \R</math>
+===Examples with repeated real roots===
+Consider the differential equation:
+<math>y'' = 6y' - 9y</math>
+First, we bring everything to one side, to get:
+<math>y'' - 6y' + 9y = 0</math>
+The characteristic polynomial is:
+<math>t^2 - 6t + 9</math>
+This factors as <math>(t - 3)^2</math>, so it has a root 3 with multiplicity 2. The general solution is thus:
+<math>y = (C_1 + C_2x)e^{3x}, \qquad C_1,C_2 \in \R</math>
+===Examples with real and complex roots without repetitions===
+Consider the differential equation:
+<math>y''' = y</math>
+First, we bring everything to one side, to get:
+<math>y''' - y = 0</math>
+The characteristic polynomial is:
+<math>t^3 - 1</math>
+The only real root is 1, corresponding to a solution basis element <math>e^x</math>. The complex roots are the roots of <math>t^2 + t + 1</math>, and they are complex conjugates:
+<math>\frac{-1 \pm \sqrt{3}i}{2}</math>
+The basis solutions for this pair are thus <math>e^{-x/2}\cos(\sqrt{3} x/2)</math> and <math>e^{-x/2}\sin(\sqrt{3} x/2)</math>. The general solution is thus:
+<math>y = C_1e^x + C_2e^{-x/2}\cos(\sqrt{3} x/2) + C_3e^{-x/2}\sin(\sqrt{3}x/2)</math>