**Examples:**

The general description of higher order linear differential equations is

$$ \begin{equation} \color{red}{a_n(x)\frac{d^n y}{dx^n} + a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}} + \dots + a_1(x)\frac{dy}{dx}+a_0(x)y = g(x)} \end{equation} $$Homogeneous DE, which has zero member $g(x)$ on the right side, is associated with non-homogeneous DE.

**Example:**

For first order DE we had only one pair of initial values. For higher order DE, initial conditions have to pass given point $x_0$ and fulfill prescribed values $y(x_0),\ y'(x_0),\ y''(x_0),\ ...$

$$ \begin{aligned} y(x_0) &= y_0\\ y'(x_0)&=y_1\\ y''(x_0) &= y_2\\ &\vdots\\ y^{n-1}(x_0) &= y_{n-1} \end{aligned} $$If all coefficients $a_n(x)$ and $g(x)$ are continuous functions on an interval $I$ containing $x_0$ and $a_n(x) \neq 0$ for every $x$ on interval, then linear DE has one and only one solution $y=y(x)$, which satisfies set of initial conditions $y(x_0) = y,\ y'(x_0) = y_1,\ \dots ,\ y^{n-1}(x_0) = y_{n-1}$.

In many cases, knowing that a unique solution exists is more important than the solution itself.

Differential equation $x^2y'' - 2xy' + 2y = 6$ has a solution $y=cx^2 +x +3$ for initial values $y(0) = 3, y'(0) = 1$.

Let us observe $a_n(x)$, i.e. $a_2(x)$. Because $a_2(x)$ is zero for $x_0 = 0$, the solution is not unique for given initial values. We can test, that every $c$ satisfies solution. There is no unique solution.

Differential equation $(x-2)y'' + 3y = x$ with initial values $y(0) = 0, y'(0) = 1$.

We are able to find an interval $(-\infty; 2)$ where $a_2(x) \neq 0$. Since $x_0$ belongs to that interval where $a_2(x) \neq 0$, there exists unique solution then.

Find the largest interval where differential equation $(t^2 -1)y'' + 3ty' + \cos t y = t$ with $y(0) = 4, y'(0) = 5$ is guaranteed to have a unique solution.

We have to use *theorem of unique solution*. The theorem states that all
coefficients $a_0(x),\ a_1(x),\ ...,\ a_n(x)$ have to be continuous on given interval. If $a_n(x) = 0$ at
a point within the interval, then continuity is broken.

Note: there is an analogy with equation $yx = 1 \implies y = 1/x$. If $x=0$ the value is not defined and the function is not continuous.

In our case for $x_0 = 0$ there exists an interval $\underline{\underline{(-1; 1)}}$ where $a_0(x),\ a_1(x),\ a_2(x)$ are continuous and $a_n(x) \neq 0$ for any $t$ from the interval.

Let $y_1,\ y_2,\ ... ,\ y_k$ be solution of the homogeneous $n$-th order DE. Then the linear combination

$$ y = c_1y_1(x) + c_2 y_2(x) + \dots + c_k y_k(x) $$with arbitrary constants $c_n$ is also solution.

If $e^{-2x}$ is a solution of given DE $y'+2y'=0$, then $ce^{-2x}$ is also solution.

If $y_1 = x^2$ and $y_2 = e^x$ are both solution of given DE, then $y = c_1x^2 + c_2e^x$ is also solution.

A set of functions $f_1(x)$, $f_2(x)$, ..., $f_n(x)$ is **linearly dependent**
on an interval $I$ if we can find constants $c_1$, $c_2$, ..., $c_n$ (not all
set to zero) such, that

for every $x$ in the interval. If the set is not linearly dependent on the interval, it is said to be linearly independent.

Can not $\implies f_1(x)$ and $f_2(x)$ are linearly independent.

If $c_2 = 0$, then $c_1$ can be nonzero and $0=0$ for every $x$ on interval. Thus $f_1(x)$ and $f_2(x)$ are linearly dependent.

When solving DE, **we are interested into linearly independent
functions.** There is a mechanism how to resolve whether given functions
are independent:

The Wronskian is a determinant, which consists of a set of functions $f_1(x),\ f_2(x),\ ...,\ f_n(x)$ which passes at least $n-1$ derivatives.

$$ W(f_1, f_2, \dots, f_n) = \begin{vmatrix} f_1(x) & f_2(x) & \dots & f_n(x)\\ f_1'(x) & f_2'(x) & \dots & f_n'(x)\\ \vdots & \vdots & & \vdots\\ f_1^{n-1}(x) & f_2^{(n-2)}(x) & \dots & f_n^{(n-1)}(x) \end{vmatrix} $$The set of functions $f_1, \ f_2, \ \dots, \ f_n$ is linearly independent if $W(f_1, \ f_2, \ \dots, \ f_n) \neq 0$ for every $x$ in the interval.

Test whether $f_1(x) = e^{3x}$ and $f_2(x)=e^{-3x}$ are linearly independent functions.

$$ W(f_1, f_2) = \begin{vmatrix} e^{3x} & e^{-3x} \\ 3e^{3x} & -3e^{-3x} \end{vmatrix} = e^{3x}(-3e^{-3x}) - e^{-3x}\cdot 3e^{3x} = -3 -3 = -6 $$$W \neq0 \implies$ the two functions are linearly independent.

The homogeneous DE of order $n$ has a fundamental set of linearly independent solutions $y_1,\ y_2,\ \dots,\ y_n$.

**General solution** is then linear combination of solutions from the set

Above function $\boldsymbol {y_c}$ is solution of homogeneous DE and is also called
**complementary function**. The higher is the degree of DE,
the more functions is involved. For example, if DE is of **2nd**
order, then $y_c(x)$ consists of **two independent
solutions.**

The nonhomogeneous DE has a solution

$$ y(x) = y_c(x) + y_p(x), $$where $y_c(x)$ is a solution found for associated homogeneous DE (above) and $y_p(x)$ is a particular solution of the nonhomogeneous DE.

$y'' + 4y'+4y = 4x^2 + 6e^x$ is nonhomogeneous DE.

Associated homogeneous DE is $y'' + 4y'+4y = 0$, which has a solution $y_c(x) = (c_1 + c_2x)e^{-2x}$.

The general solution $y(x) = y_c(x) + y_p(x)$ involves solution of associated DE and also particular solution for given nonhomogeneous DE (see member $4x^2 + 6e^x$; it will be explained in other chapter where $y_p(x)$ comes from):

$$ \begin{align} y = &(c_1 +c_2x)e^{-2x} + \hskip1em &\text{complementary function} \nonumber \\ & + x^2 -2x +\frac 3 2 + \frac 2 3 e^x \hskip1em &\text{particular solution} \nonumber \end{align} $$