# Higher order linear differential equations

## Cauchy-Euler equation

$$\color{red}{x^2}\frac{\color{red}{d^2}y}{dx^2} - 2\color{red}{x}\frac{\color{red}{d}y}{dx} - 4y = 0\\ 4\color{red}{x^2}y\color{red}{''} + 8\color{red}{x}y\color{red}{'} + y = 0\\$$

are examples of Cauchy-Euler differential equations. A linear differential equation of the form

$$a_n x^n \frac{d^n y}{dx^n} + a_{n-1} x^{n-1} \frac{d^{n-1} y}{dx^{n-1}} + \dots + a_1 x \frac{dy}{dx} + a_0y = g(x),$$

where the coefficients $a_n, a_{n-1}, \dots, a_0$ are constants, is known as a Cauchy-Euler equation. The important observation is that coefficient $x^k$ matches the order of differentiation.

Cauchy-Euler differential equations have solutions in format $\boldsymbol{y(x) = x^m}$ and are easily solved. The solution is quite straight-forward. We lay expected solution $y(x) = x^m$ and feed the solution with its derivatives $y'(x)$ and $y''(x)$ into DE. Solving leads to separating $x^m$ and auxiliary equation (see the examples below). The roots $m$ have to be determined from auxiliary equation.

Once we have roots $m$, we have to deal with 3 possibilities:

#### The roots are distinct, do not repeat

Solution is in the form $y(x) = c_1 x^{m_1} + c_2 x^{m_2}$.

#### Roots repeat

E.g. $m_1 = 2$, $m_2 = 2$. The solutions can not be linearly dependent. The Reduction of order method determines from the first solution $y_1(x) = x^{m}$ the second solution as $y_2(x) = x^{m} \log x$.

#### Roots are complex numbers

It is an analogy to the above. The imaginary roots come in conjugate pairs and solution can be written as $y(x) = c_1 x^{\alpha - i \beta} + c_2 x^{\alpha + i \beta}$. However we often prefer to work with expression without complex numbers. The other form is $y = C_1 x^{\alpha} \cos(\beta \log x) + C_2 x^{\alpha} \sin(\beta \log x)$.

Note: although the text describes mainly second order differential equation, all the ideas can be easily extended to analogous differential equation of any order.

### Example

Solve differential equation $x^2 y'' - 3xy' + 4y = 0$ with initial values $y(1) = 1$ and $y'(1) = 2$.

We assume that the solution is in the form $y(x) = x^m$. Then

$$\begin{eqnarray} y(x) &&=&& x^m, \nonumber \\ y'(x) &&=&& m x^{m-1}, \nonumber \\ y''(x) &&=&& m (m-1) x^{m-2}.\nonumber \end{eqnarray}$$

Let us use the solution within given differential equation:

$$\begin{eqnarray} \color{red}{x^2} \cdot m (m-1) \color{red}{x^{m-2}} - 3\color{red}{x} \cdot m\color{red}{x^{m-1}} + 4 \cdot \color{red}{x^m} &&=&& 0 \nonumber \\ \color{red}{x^m} ((m^2-m) - 3m + 4) &&=&& 0 \nonumber\\ x^m (m^2- 4m + 4) &&=&& 0\nonumber\\ m &&=&& \{2, 2\}\nonumber\\ \end{eqnarray}$$

The solution of differential equation is

$$\begin{equation} \underline{y_c(x) = c_1 x^2 + c_2 x^2 \log x} \label{co:ref1}. \end{equation}$$

In order to find values of $c_1$ and $c_2$ according to initial values, we need also $y_c'$:

$$\begin{eqnarray} y_c'(x) &&=&& 2 c_1 x + 2 c_2 x \log x + c_2 x^2 \frac{1}{x} = \nonumber \\ &&=&& 2c_1 x + 2 c_2 x \log x + c_2 x. \label{co:ref2} \end{eqnarray}$$

For $y(1) = 1$ and $y'(1) = 2$ we get from $(\ref{co:ref1})$ and $(\ref{co:ref2})$ $c_1 = 1$, $c_2 = 0$. The solution is then

$$\underline{\underline{y(x) = x^2}}.$$

### Example

Solve differential equation $x^2y'' - 2xy' + 2y = x^3, \hskip1em x>0$

Because the differential equation is in the form of Cauchy-Euler DE we assume that the solution is in the form of $y(x)=x^m$. Then

$$\begin{eqnarray} y(x) &&=&& x^m, \nonumber \\ y'(x) &&=&& m x^{m-1}, \nonumber \\ y''(x) &&=&& m (m-1) x^{m-2}.\nonumber \end{eqnarray}$$

Let us use the solution within associated homogeneous differential equation to obtain complementary function $y_c(x)$:

$$\begin{eqnarray} \color{red}{x^2} \cdot m (m-1) \color{red}{x^{m-2}} - 2\color{red}{x} \cdot m\color{red}{x^{m-1}} + 2 \cdot \color{red}{x^m} &&=&& 0 \nonumber \\ \color{red}{x^m} (m(m-1) - 2m + 2) &&=&& 0 \nonumber\\ x^m (m^2- 3m + 2) &&=&& 0\nonumber\\ x^m (m- 2)(m - 1) &&=&& 0\nonumber\\ m &&=&& \{1, 2\}\nonumber\\ \end{eqnarray}$$

The complementary function $y_c(x)$ (i.e. the solution of associated homogeneous differential equation) is

$$\begin{equation} \underline{y_c(x) = c_1 x + c_2 x^2} \label{co:ref3}. \end{equation}$$

We have $y_c$ but we have no $y_p$, so the solution is still incomplete. The differential equation has no constant coefficients, so we can not use method of undeterminated coefficients. We have to employ method of variation of parameters to determine particular solution $y_p$.

Let us take $y_1(x)$ and $y_2(x)$ from $y_c(x)$ and form particular solution $y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)$:

$$y_1(x) = x, \hskip1em y_2(x) = x^2 \nonumber \\ y_1'(x) = 1, \hskip1em y_2'(x) = 2x \nonumber \\$$

We have to find $u_1'(x)$ and $u_2'(x)$ as method of variation of parameters requires ($f(x) = x^3 / x^2 = x$ for differential equation in standard form):

$$\begin{eqnarray} u_1'(x) &&=&& \frac{W_1}{W} = -\frac{x^2 x}{2x^2 - x^2} = -\frac{x^3}{x^2} = -x, \nonumber\\ u_2'(x) &&=&& \frac{W_2}{W} = \frac{x\cdot x}{x^2} = 1 \nonumber \end{eqnarray}$$

After integrating $u_1'(x)$ and $u_2'(x)$ we may form particular and general solution:

$$\begin{array}{l} \underline{y_p(x) = -\frac{x^2}{2}x + x\cdot x^2 = \frac{1}{2}x^3}\\ \underline{\underline{y(x) = c_1 x + c_2 x^2 + \frac{1}{2}x^3}}. \end{array}$$

### Example

Solve differential equation $x^2y''+4xy'-4y=3$

First, let us solve associated homogeneous DE $x^2y''+4xy'-4y=0$ for $y_c$, then particular solution $y_p$.

We expect the solution in a form of $y=x^m$, then $y'=mx^{m-1},\ y''=m(m-1)x^{m-2}$.

\begin{align} x^2\ m(m-1)x^{m-2} + 4x\ mx^{m-1} - 4\ x^m &= 0 \nonumber \\ x^m[m(m-1) +4m -4] &= 0 \nonumber \\ x^m(m^2+3m-4) = 0\implies m &= \{-4, 1\} \nonumber \\ \end{align}

The complementary function is $y_c = c_1x+c_2x^{-4}$ and particular solution $y_p = A$ has to be determined and it is pretty obvious that $A=-3/4$ (test yourself that $y_p = -3/4$ satisfies given DE).

$$\underline{\underline{y = c_1x + c_2x^{-4} - \frac 3 4}}$$

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