Examples:
$$ \left. \begin{aligned} y'' + 3y' + y &= 0\\ y''' + 2y'' - y' - 2y &= 0\\ x^2y'' + xy' - y &= 0 \end{aligned} \right\} \hskip1em \text{homogeneous linear DE}\\ $$ $$ \left. \begin{aligned} y'' + 3y' + y &= \color{blue}{4}\\ y''' + 2y'' - y' - 2y &= \color{blue}{12e^x}\\ x^2y'' + xy' - y &= \color{blue}{1} \end{aligned} \right\} \hskip1em \color{blue}{\text{non}}\text{homogeneous linear DE}\\ $$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:
$$ \begin{aligned} y(1) &= 7\\ y'(1)&=0\\ y''(1) &= 0 \end{aligned} $$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
$$ c_1f_1(x) + c_2 f_2(x) + \dots + c_nf_n(x) = 0 $$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
$$ y_c(x) = c_1 y_1(x) + c_2y_2(x) + \dots + c_n y_n(x) $$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):
$$ \begin{align} y = &(c_1 +c_2x)e^{-2x} + \hskip1em &\text{complementary function} \nonumber \\ & + x^2 -2x +\frac 3 2 + \frac 2 3 e^x \hskip1em &\text{particular solution} \nonumber \end{align} $$