Two functions \(y_{1},y_{2}\) defined on an open interval \(I\) are linearly independent on \(I\) if neither is a constant multiple of the other on \(I\). Otherwise they are said to be linearly dependent on \(I\).

For example, \(y_1=e^x\) and \(y_2=e^{1+x}\) are linearly dependent because \(y_2=ey_1\). However, for \(x>0\), \(y_1=\ln {(x)}\) and \(y_2=\frac {1}{\sqrt {x}}\) are linearly independent since neither is a scalar multiple of the other, just consider that \(\ln {(x)}\) is sometimes negative.

A second order differential equation that can be written in the form \(y''+py'+qy=f\) where \(p,q,f\) are functions of the independent input variable is linear. If \(p,q\) are constant then we call the linear equation a constant coefficient equation.

It is linear and homogeneous if \(f=0\); otherwise it is linear and nonhomogeneous. We begin by focusing on the homogeneous case.

Theorem: Suppose \(p,q\) are continuous functions on an open interval \(I\), \(x_{0}\) is in \(I\) and \(k_{0},k_{1}\) are arbitrary real numbers. Then the IVP \(y''+py'+qy=0\), \(y(x_{0})=k_{0}\), \(y'(x_{0})=k_{1}\), has a unique solution on \(I\).

(We skip the proof.)

Suppose \(y_{1}\) and \(y_{2}\) are solutions to \(y''+py'+qy=0\) on an open interval \(I\) on which \(p,q\) are continuous. Then any linear combination \(y=c_{1}y_{1}+c_{2}y_{2}\) is also a solution on \(I\).

The pair \(\{y_{1},y_{2}\}\) is called a fundamental set of solutions on open interval \(I\) if all solutions to \(y''+py'+qy=0\) can be expressed as \(y=c_{1}y_{1}+c_{2}y_{2}\) for an appropriate choice of constants.
When \(\{y_{1},y_{2}\}\) is a fundamental set of solutions on \((a,b)\), \(y=c_{1}y_{1}+c_{2}y_{2}\) is said to be the general solution to \(y''+py'+qy=0\) on \(I\).

Theorem: Suppose \(p,q\) are continuous on an open interval \(I\). Then a set \(\{y_{1},y_{2}\}\) of solutions to \(y''+py'+qy=0\) on \(I\) is a fundamental set of solutions on \(I\) if and only if \(y_{1},y_{2}\) are linearly independent on \(I\).

(We skip the proof.)

Let \(y_1,y_2\) be functions. Define the Wronskian of set \(\{y_{1},y_{2}\}\) as the function \(W=y_{1}y_{2}'-y_{1}'y_{2}\), which can be written as a matrix determinant:

So \(W=y_{1}y_{2}'-y_{1}'y_{2}\). By the Product Rule, \(W'=y_{1}'y_{2}'+y_{1}y_{2}''-y_{1}''y_{2}-y_{1}'y_{2}'=\) \(y_{1}y_{2}''-y_{1}''y_{2}\).

Using that \(y_{1}''=-p y_{1}'-qy_{1}\) and \(y_{2}''=-p y_{2}'-qy_{2}\), assuming that \(y_1\) and \(y_2\) are solutions to \(y''+py'+qy=0\) on some open interval \(I\),

\(W'=y_{1}(-py_{2}'-qy_{2})-(-py_{1}'-qy_{1})y_{2}\) \(=-py_{1}y_{2}'+py_{1}'y_{2}\) \(=-p W\), a separable 1st order de!

Therefore, we have Abel’s formula for the Wronskian:

The Wronskian \(W\) of two solutions \(y_1,y_2\) to \(y''+py'+qy=0\) satisfies \(W=Ce^{-P}\) where \(P'=p\).

One important observation: If \(W(x_{0})=0\) then \(W=0\) (as \(C\) above would have to be \(0\)). If \(W(x_{0})\neq 0\) then \(W\) is NEVER \(0\).

Theorem: Suppose \(p,q\) are continuous on an open interval \(I\) and \(\{y_{1},y_{2}\}\) are solutions to \(y''+py'+qy=0\) on \(I\). Then \(y_{1},y_{2}\) are linearly independent on \(I\) if and only if the Wronskian \(W=y_{1}y_{2}'-y_{1}'y_{2}\) has no zeros on \(I\).

We will prove the forward direction (by contrapositive) and leave the other as an exercise left for you:

Suppose \(W=0\) at some point on \(I\). Then by Abel’s formula, \(W=0\) at every point on \(I\). If \(y_1\) is \(0\) then the functions \(y_1\) and \(y_2\) are linearly dependent on \(I\). Suppose \(y_1\neq 0\); since it is differentiable on \(I\) and thus continuous on \(I\), \(y_1\) is never \(0\) on at least some open interval \(J\) contained in \(I\). Then on \(J\), \((y_{2}/y_{1})\) is defined and \(\displaystyle \left (\frac {y_{2}}{y_{1}}\right )'=\frac {y_{1}y_{2}'-y_{1}'y_{2}}{y_{1}^{2}}=\frac {W}{y_{1}^{2}}=0\).

That means that on \(J\), \(y_2=Cy_1\) for some constant \(C\).

We’d like to show this is the case on \(I\). Let \(Y=Cy_{1}-y_{2}\). Since \(Y\) is a linear combination of \(y_{1}\) and \(y_{2}\), it is a solution of the IVP \(y''+py'+qy=0\), with \(y(x_{0})=y'(x_{0})=0\) for some \(x_{0}\) in \(J\).

Since the trivial solution \(Y=0\) works and IVP has a unique solution, it must be the case that \(Y=0\) on \(I\) and hence \(y_{2}=Cy_{1}\) on \(I\).

Hence the functions are linearly dependent!

Example:

One can show that both \(y_1=x\) and \(\displaystyle y_2=\frac {1}{x^2}\) are solutions to \(\displaystyle y''+\frac {2}{x}y'-\frac {2}{x^2}y=0\) on \((0,\infty )\). Let’s use the Wronskian to justify that these are linearly independent on \((0,\infty )\) and thereby determine the general solution to this de.

So the Wronskian is indeed never zero on \((0,\infty )\). Another way to see that the Wronskian is never zero, is to observe that \(y_1\neq 0\) on \((0,\infty )\) and that \(\frac {y_2}{y_1}=x^{-3}\) is non-constant; hence \(\displaystyle \left (\frac {y_2}{y_1}\right )'=\frac {W}{y_1^2}\) is non-zero. So \(W\neq 0\), and by Abel’s formula, if \(W\) is not identically zero on \((0,\infty )\) then it is NEVER zero on the \((0,\infty )\).

Either way we see that \(y_1=x\) and \(y_2=x^{-2}\) are indeed linearly independent, and form a fundamental set of solutions to this second order linear homogeneous differential equation. So the general solution to this de is