# 7.1 Second-order linear equations  (Page 3/15)

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It turns out that to find the general solution to a second-order differential equation, we must find two linearly independent solutions. We define that terminology here.

## Definition

A set of functions ${f}_{1}\left(x\right),{f}_{2}\left(x\text{),…,}{f}_{n}\left(x\right)$ is said to be linearly dependent    if there are constants ${c}_{1},{c}_{2}\text{,…}{c}_{n},$ not all zero, such that ${c}_{1}{f}_{1}\left(x\right)+{c}_{2}{f}_{2}\left(x\right)+\text{⋯}+{c}_{n}{f}_{n}\left(x\right)=0$ for all x over the interval of interest. A set of functions that is not linearly dependent is said to be linearly independent    .

In this chapter, we usually test sets of only two functions for linear independence, which allows us to simplify this definition. From a practical perspective, we see that two functions are linearly dependent if either one of them is identically zero or if they are constant multiples of each other.

First we show that if the functions meet the conditions given previously, then they are linearly dependent. If one of the functions is identically zero—say, ${f}_{2}\left(x\right)\equiv 0$ —then choose ${c}_{1}=0$ and ${c}_{2}=1,$ and the condition for linear dependence is satisfied. If, on the other hand, neither ${f}_{1}\left(x\right)$ nor ${f}_{2}\left(x\right)$ is identically zero, but ${f}_{1}\left(x\right)=C{f}_{2}\left(x\right)$ for some constant $C,$ then choose ${c}_{1}=\frac{1}{C}$ and ${c}_{2}=-1,$ and again, the condition is satisfied.

Next, we show that if two functions are linearly dependent, then either one is identically zero or they are constant multiples of one another. Assume ${f}_{1}\left(x\right)$ and ${f}_{2}\left(x\right)$ are linearly independent. Then, there are constants, ${c}_{1}$ and ${c}_{2},$ not both zero, such that

${c}_{1}{f}_{1}\left(x\right)+{c}_{2}{f}_{2}\left(x\right)=0$

for all x over the interval of interest. Then,

${c}_{1}{f}_{1}\left(x\right)=\text{−}{c}_{2}{f}_{2}\left(x\right).$

Now, since we stated that ${c}_{1}$ and ${c}_{2}$ can’t both be zero, assume ${c}_{2}\ne 0.$ Then, there are two cases: either ${c}_{1}=0$ or ${c}_{1}\ne 0.$ If ${c}_{1}=0,$ then

$\begin{array}{c}0=\text{−}{c}_{2}{f}_{2}\left(x\right)\hfill \\ 0={f}_{2}\left(x\right),\hfill \end{array}$

so one of the functions is identically zero. Now suppose ${c}_{1}\ne 0.$ Then,

${f}_{1}\left(x\right)=\left(-\frac{{c}_{2}}{{c}_{1}}\right){f}_{2}\left(x\right)$

and we see that the functions are constant multiples of one another.

## Linear dependence of two functions

Two functions, ${f}_{1}\left(x\right)$ and ${f}_{2}\left(x\right),$ are said to be linearly dependent if either one of them is identically zero or if ${f}_{1}\left(x\right)=C{f}_{2}\left(x\right)$ for some constant C and for all x over the interval of interest. Functions that are not linearly dependent are said to be linearly independent .

## Testing for linear dependence

Determine whether the following pairs of functions are linearly dependent or linearly independent.

1. ${f}_{1}\left(x\right)={x}^{2},$ ${f}_{2}\left(x\right)=5{x}^{2}$
2. ${f}_{1}\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ ${f}_{2}\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$
3. ${f}_{1}\left(x\right)={e}^{3x},$ ${f}_{2}\left(x\right)={e}^{-3x}$
4. ${f}_{1}\left(x\right)=3x,$ ${f}_{2}\left(x\right)=3x+1$
1. ${f}_{2}\left(x\right)=5{f}_{1}\left(x\right),$ so the functions are linearly dependent.
2. There is no constant C such that ${f}_{1}\left(x\right)=C{f}_{2}\left(x\right),$ so the functions are linearly independent.
3. There is no constant C such that ${f}_{1}\left(x\right)=C{f}_{2}\left(x\right),$ so the functions are linearly independent. Don’t get confused by the fact that the exponents are constant multiples of each other. With two exponential functions, unless the exponents are equal, the functions are linearly independent.
4. There is no constant C such that ${f}_{1}\left(x\right)=C{f}_{2}\left(x\right),$ so the functions are linearly independent.

Determine whether the following pairs of functions are linearly dependent or linearly independent: ${f}_{1}\left(x\right)={e}^{x},$ ${f}_{2}\left(x\right)=3{e}^{3x}.$

Linearly independent

If we are able to find two linearly independent solutions to a second-order differential equation, then we can combine them to find the general solution. This result is formally stated in the following theorem.

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