# Ordinary Differential Equations/Linear Equations

## Linear Equations and Linear OperatorsEdit

A general linear equation is of the form

${\displaystyle P_{0}(x)y^{(n)}+P_{1}(x)y^{(n-1)}+...+P_{n}y=Q(x)}$ ,

where we will assume that the coefficient functions ${\displaystyle P_{i}(x)}$  and Q(x) are continuous functions in an interval [a,b], and that ${\displaystyle P_{0}(x)\neq 0}$  for all ${\displaystyle x\in [a,b]}$ . The existence theorem proves that a unique solution exists that passes through the point ${\displaystyle (x_{0},y_{0})}$  where ${\displaystyle x\in [a,b]}$ , with continuous derivatives up to the n-1 order, and satisfies initial conditions for each of those derivatives at ${\displaystyle x_{0}}$ .

One can also write this as

${\displaystyle L(y)=(P_{0}{\frac {d^{n}}{dx^{n}}}+P_{1}{\frac {d^{n-1}}{dx^{n-1}}}+...+P_{n})y=Q(x)}$

where L is called a linear differential operator of order n.

A differential equation of the form L(y)=0 with the same linear differential operator as above is called a homogeneous equation corresponding to the above equation, and the reduced equation of the above equation.

We will now prove some properties about the linear differential operator.

The linear differential operator:

${\displaystyle L(C_{1}y_{1}+C_{2}y_{2})=C_{1}L(y_{1})+C_{2}L(y_{2})}$  which is true because differentiation is linear. Thus, we can make the two following statements:

• If ${\displaystyle y_{1}}$  and ${\displaystyle y_{2}}$  are two solutions of the homogeneous equation, then ${\displaystyle C_{1}y_{1}+C_{2}y_{2}}$  is also a solution.
• If y is a solution to the homogeneous equation L(y)=0, and ${\displaystyle y_{0}}$  is a solution to the equation L(y)=Q(x), then ${\displaystyle y+y_{0}}$  is also a solution to the equation L(y)=Q(x).

## WronskianEdit

Let L(y)=0 be a homogeneous equation of degree n, and let ${\displaystyle y_{1},y_{2},...,y_{n}}$  be linearly independent solutions of this equation.

The general solution is then ${\displaystyle y=C_{1}y_{1}+C_{2}y_{2}+...+C_{n}y_{n}}$ .

Suppose instead that the solutions are not linearly independent. Then there exists values for ${\displaystyle C_{1},C_{2},...,C_{n}}$  not all zero such that ${\displaystyle C_{1}y_{1}+C_{2}y_{2}+...+C_{n}y_{n}}$ =0.

Then the following equations also hold true:

${\displaystyle C_{1}y_{1}'+C_{2}y_{2}'+...+C_{n}y_{n}'=0}$
${\displaystyle C_{1}y_{1}''+C_{2}y_{2}''+...+C_{n}y_{n}''=0}$
...
${\displaystyle C_{1}y_{1}^{(n-1)}+C_{2}y_{2}^{(n-1)}+...+C_{n}y_{n}^{(n-1)}=0}$

When there is to be a non-trivial solution to this system of homogeneous linear equations, the column vectors corresponding to each coefficient are linearly dependent. This is equivalent to saying that the following determinant, called the Wronskian is 0:

${\displaystyle {\begin{Vmatrix}y_{1}&y_{2}&...&y_{n}\\y_{1}'&y_{2}'&...&y_{n}'\\y_{1}''&y_{2}''&...&y_{n}''\\...&...&...&...\\y_{1}^{(n-1)}&y_{2}^{(n-1)}&...&y_{n}^{(n-1)}\\\end{Vmatrix}}=0}$

Thus, if the solutions are dependent, then their Wronskian is equal to 0, and conversely, and conversely if the Wronskian is equal to 0, then the columns of that matrix must be linearly dependent, indicating that the solutions are linearly dependent.