Higher order linear differential equations

Homogeneous DE with constant coefficients

Example: we want to solve $y'''+2y''-y'-2y = 0$ . We will show, that it leads to solution of polynom $m^3 + 2 m^2 -m -2 = 0$ and the solution involves members $y= e^{mx}$ .

We have learnt from linear DE of first order that the solution contains member $e^{mx}$ . Without going into details, let us say that the possible solution of differential equation

$\begin{equation} a_n y^{(n)} + a_{n-1} y^{n-1} + \dots + a_1y' + a_0 y = 0\label{ref:DE} \end{equation}$

$\begin{equation} y = e^{mx}. \label{ref:solution} \end{equation}$

For what value of $m$ will $(\ref{ref:solution})$ be a solution of $(\ref{ref:DE})$ ?

$\begin{equation} a_n \frac{d^n}{dx^n}e^{mx} + a_{n-1}\frac{d^{n-1}}{dx^{n-1}}e^{mx} + \dots a_{1}\frac{d}{dx}e^{mx} + a_0 e^{mx} = 0. \label{ref:eq3} \end{equation}$

Since $k$ -th derivative of $e^{mx} = m^k e^{mx}$ , we may rewrite $(\ref{ref:eq3})$ to

$\begin{equation} a_n m^n e^{mx} + a_{n-1}m^{n-1}e^{mx} + \dots + a_1me^{mx} + a_0 e^{mx} = 0 \label{ref:eq4} \end{equation}$

Let us divide $(\ref{ref:eq4})$ by $e^{mx}$ to obtain auxiliary (characteristic) equation:

$\begin{equation} \color{red}{{a_nm^n + a^{n-1}m^{n-1} + \dots + a_1m + a_0 = 0}}. \end{equation}$

The task diminished to an algebra task of finding roots $m$ . We have to deal with the facts that

We can find roots of quadratic equation $ax^2 + bx + c = 0$ , but it is not a simple task to find roots of higher order polynomial equations.
There are 3 possibilities what kind of roots might occur:

Roots are distinct and real (e.g. $m_1 = 1$ , $m_2 = 3$ )
Some of the roots repeat (e.g. $m_1 = 2$ and $m_2 = 2$ )
Roots are imaginary (e.g. $m_1 = 11 + 2i$ , $m_2 = 11 - 2i$ )
The imaginary roots must occur in conjugate pairs $\alpha \pm i\beta$

A. Roots of auxiliary equation are real and distinct

The solution consists of functions $y_1 = e^{m_1 x},\ y_2 = e^{m_2 x},\ \dots,\ y_n = e^{m_n x}$ or

$y_c = c_1 e^{m_1 x} + c_2 e^{m_2x} + \dots + c_n e^{m_n x}.$

Because the roots do not repeat, is is given that above terms are linearly independent.

Example

Find the general solution of $y'''+2y''-y'-2y = 0.$

We construct the auxiliary equation:

$m^3 + 2m^2 -m-2 = 0 \implies m = \{-2, 1, -1 \}.$

The solution is

$\underline{\underline{y_c = c_1 e^{-2x} + c_2e^{x} + c_3e^{-x}}}.$

Example

Find particular solution $y(x)$ of $y'' - 3y'+2y = 0$ for which $y_c(0) = 1$ and $y_c'(0) = 0$ .

Auxiliary form is

$\begin{align} & m^2 - 3m +2 = 0 \implies m = \{1, 2\}. \nonumber\\ & y_c(x) = c_1e^x + c_2 e^{2x}\nonumber\\ & y_c'(x) = c_1e^x + 2 c_2 e^{2x}\nonumber\\ \end{align}$

Substitute initial values:

$\left. \begin{aligned} & 1 = c_1 e^0 + c_2 e^0\\ & 0 = c_1 e^0 + 2c_2 e^0 \end{aligned} \right\} \begin{aligned} &c_1=2\\ &c_2 = -1 \end{aligned}$

Substituting constants into complementaty function $y_c$ we have particular solution for given initial values:

$\underline{\underline{y_c = 2e^{x} - e^{2x}}}.$

B. Some roots are multiple

Even if the roots repeat, the members of solution can not repeat. Because we expect only functions, which are linearly independent. It can be shown (reduction of order method), that if roots are $m_1 = m_2$ , the solution is $e^{m_1 x} + xe^{m_2 x}$ .

In general it can be shown that if auxiliary equation has a root $m=a$ which repeats $n$ times, then the general solution is

$y_c = (c_1 + c_2x + c_3 x^2 + \dots c_nx^n) e^{ax}.$

Example

Find the general solution of $y(x)$ of $y^{(4)} - 3y''+2y' = 0$ .

$\begin{aligned} &m^4 -3m^2 + 2m = 0 \implies m =\{0, 1, 1, -2\}\\ &y_c = c_1e^{0x} + c_2 e^x +c_3 x e^x + c_4 e^{-2x}\\ &\underline{\underline{y_c = c_1 + c_2 e^x +c_3 x e^x + c_4 e^{-2x}}}\\ \end{aligned}$

C. Roots are imaginary

The imaginary root must occur in conjugate pairs: if $\alpha +\beta$ is one root, another root must be $\alpha -i\beta$ .

If roots are complex $\alpha \pm i\beta$ then the solution is

$y_c = c_1 e^{(\alpha -i\beta)x} + c_2e^{(\alpha +i\beta)x}.$

We usually do not want to work with complex numbers. The solution above can be written also as

$y_c = e^{\alpha x} (c_1 \cos \beta x + c_2 \sin \beta x).$

Example

Find the general solution of $y^{(4)} + 2y'' + y = 0$ .

$m^4 + 2m^2 + 1 = 0 \hskip1em \text{or} \hskip1em (m^2 + 1)^2 = 0 \implies m = \{i, i, -i, -i\}$

Since complex number has a form $\alpha \pm i\beta$ , we have $\beta = 1$ and the number repeats.

$\begin{aligned} & y_c = c_1e^{ix} + c_2^{-ix} + c_3e^{ix}x + c_4e^{-ix}x\\ & \underline{\underline{y_c = (c_1 + c_3)x e^{ix} +(c_2 + c_4)x e^{-ix} }}\\ & \text{or}\\ & y_c = e^{0}(c_1 \cos x + c_2 \sin x) + e^{0}x (c_3 \cos x + c_4 \sin x)\\ & \underline{\underline{y_c = (c_1 + c_3 x) \cos x+ (c_2 + c_4 x) \sin x}}\\ \end{aligned}$

Problem solving

Example

Solve $y''+4y'+4y = 0$ .

$\begin{aligned} &m^2 + 4m + 4 =0 \implies m = \{-2, -2\}\\ &\underline{\underline{y = c_1e^{-2x} + c_2xe^{-2x}}} \end{aligned}$

Example

Solve $y^{(4)} - 2y''=0$ .

$\begin{aligned} & m^4 - 2m^2 = 0\\ & (m-\sqrt{2})^2 m^2 = 0 \implies m= \{-\sqrt{2}, \sqrt{2}, 0 , 0\}\\ & y_c = c_1e^{-\sqrt{2}x} + c_2e^{\sqrt{2}x}+ c_3e^0 + c_4xe^{0} \\ & \underline{\underline{y_c = c_1e^{-\sqrt{2}x} + c_2e^{\sqrt{2}x}+ c_3 + c_4x}}\\ \end{aligned}$

Example

Solve $y''-2y'+5y = 0$ .

$\begin{aligned} &m^2 - 2m + 5 =0 \implies m = \{1 + 2i, 1-2i\}\hskip1em(\alpha = 1, \beta =2)\\ &y_c = c_1e^{(1+2i)x} + c_2 e^{(1-2i)x}\\ &\underline{\underline{y_c = e^x(c_1 \cos 2x + c_2\sin 2x)}} \end{aligned}$

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