Chapter 4

4.1

$n$th order differential equations

$$a _n (x) y ^{(n)} (x) + a _{n-1} (x) y ^{(n - 1)} (x) + \dots + a _2 (x) y\prime\prime(x) + a _1 (x) y\prime(x) + a_0(x)y(x) = f(x)$$

if $f(x) = 0$ then "homogeneous"
otherwise "non-homogeneous"

Example: $$y\prime \prime - 4y = 0$$

linear, 2nd order, homogeneous, constant coefficient

Example: $$y \prime \prime + ry\prime - 3y = 2t$$

linear, 2nd order, non-homogeneous, constant coefficient

Example: $$3x^3y\prime\prime\prime-6xy+y=10e^x$$

linear, 3rd order, non-homogeneous

Example: $$y\prime\prime\prime + 49y = 0$$

Guessed Solution:

$$y(x) = \left\{ \sin(7x), \cos(7x) \right\}$$

Suppose we have an $n$th-order linear, homogeneous DE:

$$a _n (x) y ^{(n)} (x) + a _{n-1} (x) y ^{(n - 1)} (x) + \dots + a _2 (x) y\prime\prime(x) + a _1 (x) y\prime(x) + a_0(x)y(x) = 0$$

if we find $n$ solutions to it, [unclear] consider

$$y = C_1y_1 + C_2y_2 + \dots + C_ny_n \tag{*}$$

Initial conditions: $$\begin{align} y(x_0) =& b_0 \\ y\prime(x_0) =& b_1 \\ & \vdots \\ y^{(n-1)}(x_0) =& b_{n-1} \\ \end{align}$$

Plug in ICs for $\text{(*)}$:

$$\begin{align} \text{(**)} \left\{ \begin{align} C_1y_1(x_0) & + C_2y_2(x_0) & + \dots & + C_ny_n(x_0) &= b_0& \\ C_1y_1\prime(x_0) & + C_2y_2\prime(x_0) & + \dots & + C_ny_n\prime(x_0) &= b_1& \\ C_1y_1\prime\prime(x_0) & + C_2y_2\prime\prime(x_0) & + \dots & + C_ny_n\prime\prime(x_0) &= b_2& \\ & \vdots \\ C_1y_1^{(n-1)}*(x_0) & + C_2y_2^{(n-1)}(x_0) & + \dots & + C_ny_n^{(n-1)}(x_0) &= b_n& \\ \end{align} \right\} \Leftrightarrow &\left[ \begin{array}{cccc} y_1(x_0) & y_2(x_0) & \dots & y_n(x_0) \\ y_1\prime(x_0) & y_2\prime(x_0) & \dots & y_n\prime(x_0) \\ \vdots & & & \\ y_1^{(n-1)}(x_0) & y_2^{(n-1)}(x_0) & \dots & y_n^{(n-1)}(x_0) \\ \end{array} \right] \left[ \begin{array}{c} C_1 \\ C_2 \\ \vdots \\ C_n \\ \end{array} \right] = \left[ \begin{array}{c} b_1 \\ b_2 \\ \vdots \\ b_n \\ \end{array} \right] \end{align}$$

Only valid if $W = \det(\text{(**)}) \neq 0$

($W$: "Wronskian")

For the previous Example:

$$\begin{align} y_1(x) &= \sin(7x) \\ y_2(x) &= \cos(7x) \\ \end{align}$$

$$\begin{align} \left| \begin{array}{cc} \sin(7x) & \cos(7x) \\ 7\cos(7x) & -7\sin(7x) \\ \end{array} \right| &= -7\sin^2(7x)-7\cos^2(7x) \\ &= -7\left(\sin^2(7x)\cos^2(7x)\right) \\ &= -7 \\ &\neq 0 \end{align}$$

Linear independence of a function

Definition

We say $y_1(x), y_2(x), \dots, y_n(x)$ is linearly independent on the interval $I$ if $C_1y_1(x) + C_2y_2(x) + \dots + C_nY_n(x)$ on $I$ only has the solution $C_1 = C_2 = \dots = C_n = 0$

Example: $$\begin{align} y_1(x) &= \cos^2(x) \\ y_2(x) &= 8\sin^2(x) \\ y_3(x) &= 4 \\ \end{align}$$ Is $\left\{y_1(x), y_2(x), y_3(x)\right\}$ linearly independent over $\R$? Observe: $$\begin{align} 8y_1 + y_2 &= 8\cos^2(x)+8\sin^2(x) \\ &= 8 \\ &= 2y_3(x) \\ \end{align}$$

$$\begin{align} \Rightarrow 8y_1(x) + y_2(x) - 2y_3(x) = 0 &&& \text{For all x in $\R$} \\ \Rightarrow \left\{y_1(x), y_2(x), y_3(x)\right\} &&& \text{is not linearly independent over $\R$} \end{align}$$

If $\left\{ y_1(x), y_2(x), \dots, y_n(x) \right\}$ is a collection of $n-1$ differentiable functions over $I$, and $W(x_0) \neq 0$ at any pt $x \in I$, then $\left\{ y_1(x), y_2(x), \dots, y_n(x) \right\}$ is linearly independent.

Conversely, if $y_i(x)$s are infinitely differentiable, then $W(x)= 0$ implies linear dependence.

Example: $$\begin{align} y_1(x) &= x \\ y_2(x) &= |x| \\ \end{align}$$ Is $\left\{ y_1(x), y_2(x) \right\}$ linearly independent on $\underbrace{(-1, 1)}_{I}$?

Observe: $$\begin{align} \text{for any } & x \ge 0, & y_1(x) = x = y_2(x) \Rightarrow y_1(x) - y_2(x) = 0 \\ \text{for any } & x \lt 0, & y_1(x) + y_2(x) = 0 \end{align}$$ Since constants differ, $\left\{y_1, y_2\right\}$ is independent.

4.3 - Homogeneous linear DEs with constant coefficients

$$a _n y ^{(n)} + a _{n-1} y ^{(n - 1)} + \dots + a _2 y\prime\prime + a _1 y\prime + a_0y = 0$$

$a_i$ constant

Educated guess that $e^{\lambda x}$ is the solution

Plug it in:

$$\begin{align} a_n(e^{\lambda x})^{(n)} + a_{n-1}(e^{\lambda x})^{(n - 1)} + \dots + a_{1}(e^{\lambda x})\prime + a_0e^{\lambda x} &= 0 \\ a_n \lambda^n e^{\lambda x} + a_{n-1} \lambda^{n-1} e^{\lambda x} + \dots + a_1 \lambda e^{\lambda x} + a_0 e^{\lambda x} &= 0 \\ e^{\lambda x}\left[a_n \lambda^n + a_{n-1} \lambda^{n-1} + \dots + a_1 \lambda + a_0 \right] &= 0 \\ \Leftrightarrow a_n \lambda^n + a_{n-1} \lambda^{n-1} + \dots + a_1 \lambda + a_0 &= 0 &&& \text{"Characteristic Equation"} \\ \end{align}$$

Fundamental Theorem of Algebra: The characteristic equation in principle always has $n$ roots, $\lambda_1, \lambda_2, \dots, \lambda_n$, some may repeat, some may be complex.

Simple case when $n = 2$:

$$y\prime\prime + py\prime + 2y = 0$$

Form characteristic equation:

$$\lambda^2 + p\lambda + q = 0$$

Cases:

$$\begin{align} \frac{-p \pm \sqrt{p^2-4q}}{2} \Rightarrow \begin{cases} \text{(1) } p^2-4q \gt 0 \Rightarrow \begin{cases} \lambda_1 = \frac{-p + \sqrt{p^2-4q}}{2} \\ \lambda_2 = \frac{-p - \sqrt{p^2-4q}}{2} \\ \end{cases} \\ \text{(2) } p^2-4q = 0 \Rightarrow \lambda = \frac{-p}{2} \\ \text{(3) } p^2-4q \lt 0 \Rightarrow \begin{cases} \lambda_1 = \frac{-p + i\sqrt{p^2-4q}}{2} \\ \lambda_2 = \frac{-p - i\sqrt{p^2-4q}}{2} \\ \end{cases} \end{cases} \end{align}$$

Example of case 1 (real roots):

$$\begin{align} y\prime\prime + 4y\prime + 3y &= 0 \\ \begin{cases} y(0) &= 1 \\ y\prime(0) &= 2 \\ \end{cases} \end{align}$$

$$\begin{align} \lambda^2 + 4\lambda + 3 &= 0 \\ \left(\lambda + 1 \right)\left(\lambda + 3\right) = 0 \\ \Rightarrow \lambda &= -1, 3 \\ \Rightarrow y_1(x) = e^{-1x}, y_2(x) = e^{-3x} \end{align}$$

General Solution: $$\begin{align} y(x) &= C_1e^{-x} + C_2e^{-3x} \\ y\prime(x) &= -C_1e^{-x} + -3C_2e^{-3x} \end{align}$$ Specific Solution: $$\begin{align} &\begin{cases} C_1 + C_2 &= 1\\ -C_1 - 3C_2 &= 2 \\ \end{cases} \\ & \Rightarrow -2C_2 = -1 \\ & \Rightarrow C_2 = \frac{1}{2} \\ & \Rightarrow C_1 = \frac{1}{2} \\ \end{align}$$

Example for case 2 (repeated roots): $$\begin{align} y_1(t) = e^{\lambda t} \\ y_2(t) = ? \\ \rightarrow \text{Guess:} y_2(t) = te^{^\lambda t} \end{align}$$

Check $$\begin{align} 0 &= \left(te^{\lambda t}\right)\prime\prime + p\left(te^{\lambda t}\right)\prime + qte^{\lambda t} \\ &= \left(\lambda e^{\lambda t} + t\lambda e^{\lambda t}\right)\prime + p\lambda t e^{\lambda t} + pe^{\lambda t} + qte^{\lambda t} \\ &= \lambda e^{\lambda t} + \lambda e^{\lambda t} + t\lambda^2e^{\lambda t} + p\lambda t e^{\lambda t} + pe^{\lambda t} + qte^{\lambda t} \\ &= 2\lambda e^{\lambda t} + \left(\lambda^2 + p\lambda + q\right)\left(te^{\lambda t}\right) \\ &= ??? \\ &= \left(2\lambda + p\right)\left(e^{\lambda t}\right) \\ &= 2\left(\frac{-p}{2}\right) + p \\ &= 0 \, \, \checkmark \end{align}$$ Now Check $$\begin{align} &\left| \begin{array}{cc} y_1 & y_2 \\ y_1\prime & y_2\prime \\ \end{array} \right| \\ &= \left| \begin{array}{cc} e^{\lambda t} & te^{\lambda t} \\ \lambda e^{\lambda t} & e^{\lambda t} + \lambda te^{\lambda t} \\ \end{array} \right| \\ &= e^{2\lambda t} + \cancel{\lambda te^{2\lambda t} - \lambda te^{2\lambda t}} \\ &= e^{2\lambda t} \\ &\neq 0 \, \, \checkmark \end{align}$$

Example: $$\begin{align} y\prime\prime - 6y\prime + 9y &= 0 \\ \text{IC} &= \begin{cases} y(0) &= 1 \\ y\prime(0) &= -1 \end{cases} \\ \end{align}$$ Step 1: Find general solution: $$\begin{align} 0 &= \lambda^2 - 6\lambda + 9 \\ &= \left(\lambda - 3\right)^2 \\ &\Rightarrow \lambda = 3 \\ &\Rightarrow y(t) = C_1e^{3t} + C_2te^{3t} \\ \end{align}$$ Step 2: Plug in ICs $$\begin{align} y\prime(t) = 3C_1e^{3t} + C_2\left(e^{3t}+3te^{3t}\right) \\ \begin{cases} 1 &= y(0) = C_1 \\ -1 &= y\prime(0) = 3C_1+C_2 = 3 + C_2 \Rightarrow C_2 = -4 \\ \end{cases} \end{align}$$

Case 3 (complex roots): $$\begin{align} \lambda &= \frac{-p \pm i\sqrt{4q - p^2}}{2} \\ &=\frac{-p}{2} \pm i\frac{\sqrt{4q - p^2}}{2} \\ &\Big\Downarrow \\ e^{\lambda t} &= e^{\frac{-p}{2}\cdot t \pm i\cdot\frac{\sqrt{4q - p^2}}{2}\cdot t} \\ &= e^{\frac{-p}{2}\cdot t} \pm e^{i\cdot\frac{\sqrt{4q - p^2}}{2}\cdot t} \\ &= e^{\frac{-p}{2}\cdot t} \cdot \pm \left(\cos\left({\frac{\sqrt{4q - p^2}}{2}\cdot t}\right) \pm i\sin\left({\frac{\sqrt{4q - p^2}}{2}\cdot t}\right)\right) \\ &= ??? \\ &\Big\Downarrow \\ y_1(t) &= e^{\frac{-p}{2}\cdot t} \cos\left({\frac{\sqrt{4q - p^2}}{2}\cdot t}\right) \\ y_2(t) &= e^{\frac{-p}{2}\cdot t} \sin\left({\frac{\sqrt{4q - p^2}}{2}\cdot t}\right) \\ \Rightarrow y(t) &= C_1e^{\frac{-p}{2}\cdot t} \cos\left({\frac{\sqrt{4q - p^2}}{2}\cdot t}\right) + C_2e^{\frac{-p}{2}\cdot t} \sin\left({\frac{\sqrt{4q - p^2}}{2}\cdot t}\right) \end{align}$$

Example: $$\begin{align} &\begin{cases} y\prime\prime - 4y\prime + 5y = 0 \\ y(0) = 2 \\ y\prime(0) = -3 \end{cases} \\ &\lambda^2 - 4\lambda + 5 = 0 \\ &\Rightarrow \lambda = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i \end{align}$$ General Solution: $$\begin{align} y(t) &= C_1e^{2t}\cos(t) + C_2e^{2t}\sin(t) \\ y\prime(t) &= C_1\left(2e^{2t}\cos(t) - e^{2t}\sin(t)\right) + C_1\left(2e^{2t}\sin(t) + e^{2t}\cos(t)\right) \end{align}$$ Solve IVP: $$\begin{align} 2 &= y(0) = C_1 \\ -3 &= y\prime(0) = C_1(2)+C_2(2) = 2C_1 + C_2 = 4 + C_2 = -3 \Rightarrow C_2 = -7 \\ \\ y(t) &= 2e^{2t}\cos(t) - 7e^{2t}\sin(t) \\ \end{align}$$ .

$$

Higher order ($n \gt 3$) linear, homogeneous, constant coefficient eqns

$$a_ny^{(n)} + a_{n - 1}y^{(n - 1)} + \dots + a_2y\prime\prime + a_1y\prime a_0y = 0$$

Characteristic Equation:

$$\lambda ^n a_n y^{(n)} + \lambda ^{n-1} a_{n-1}n y^{(n - 1)} + \dots + a_1\lambda + a_0 = 0$$

Roots: $\lambda_1, \dots, \lambda_n$

General Idea:

Each distinct real root $\lambda_i \Leftrightarrow y_i(t)e^{\lambda_i t}$
Each pair of distinct complex roots:

$$\begin{align} \lambda = a \pm bi \Leftrightarrow \begin{cases} y_+(t) &= e^{at}\cos(bt) \\ y_-(t) &= e^{at}\sin(bt) \\ \end{cases} \end{align}$$

If any repetition happens ($k$ times):

$$\begin{align} \begin{cases} y_1 &= e^{\lambda t}, te^{\lambda t}, \dots, t^{k-1}e^{\lambda t} \\ y_\pm &= e^{at}\cos(bt), e^{at}\sin(bt), \dots, \widetilde{y}_\pm = t^{k-1}e^{at}\cos(bt), t^{k-1}e^{at}\sin(bt) \end{cases} \end{align}$$

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