# Second Order Linear Homogeneous Differential Equations with Variable Coefficients

A linear homogeneous second order equation with variable coefficients can be written as

$y^{\prime\prime} + {a_1}\left( x \right)y' + {a_2}\left( x \right)y = 0,$

where a1(x) and a2(x) are continuous functions on the interval [a, b].

## Linear Independence of Functions. Wronskian

The functions $${y_1}\left( x \right),{y_2}\left( x \right), \ldots ,{y_n}\left( x \right)$$ are called linearly dependent on the interval $$\left[ {a,b} \right],$$ if there are constants $${\alpha _1},{\alpha _2}, \ldots ,{\alpha _n},$$ not all zero, such that for all values of $$x$$ from this interval, the identity

${\alpha _1}{y_1}\left( x \right) + {\alpha _2}{y_2}\left( x \right) + \ldots + {\alpha _n}{y_n}\left( x \right) \equiv 0$

holds. If this identity is satisfied only when $${\alpha _1} = {\alpha _2} = \ldots$$ $$= {\alpha _n} = 0,$$ then these functions $${y_1}\left( x \right), {y_2}\left( x \right), \ldots ,$$ $${y_n}\left( x \right)$$ are called linearly independent on the interval $$\left[ {a,b} \right].$$

For the case of two functions, the linear independence criterion can be written in a simpler form: The functions $${y_1}\left( x \right),$$ $${y_2}\left( x \right)$$ are linearly independent on the interval $$\left[ {a,b} \right],$$ if their quotient in this segment is not identically equal to a constant:

$\frac{{{y_1}\left( x \right)}}{{{y_2}\left( x \right)}} \ne \text{const.}$

Otherwise, when $${\frac{{{y_1}\left( x \right)}}{{{y_2}\left( x \right)}}} \equiv \text{const,}$$ these functions are linearly dependent.

Let $$n$$ functions $${y_1}\left( x \right),$$ $${y_2}\left( x \right), \ldots ,$$ $${y_n}\left( x \right)$$ have derivatives of $$\left( {n - 1} \right)$$ order. The determinant

$W\left( x \right) = {W_{{y_1},{y_2}, \ldots ,{y_n}}}\left( x \right) = \left| {\begin{array}{*{20}{c}} {{y_1}}&{{y_2}}& \ldots &{{y_n}}\\ {{y'_1}}&{{y'_2}}& \ldots &{{y'_n}}\\ \ldots & \ldots & \ldots & \ldots \\ {y_1^{\left( {n - 1} \right)}}&{y_2^{\left( {n - 1} \right)}}& \ldots &{y_n^{\left( {n - 1} \right)}} \end{array}} \right|$

is called the Wronski determinant or Wronskian for this system of functions.

### Wronskian Test.

If the system of functions $${y_1}\left( x \right),$$ $${y_2}\left( x \right), \ldots ,$$ $${y_n}\left( x \right)$$ is linearly dependent on the interval $$\left[ {a,b} \right],$$ then its Wronskian vanishes on this interval.

It follows from here that if the Wronskian is nonzero at least at one point in the interval $$\left[ {a,b} \right],$$ then the functions $${y_1}\left( x \right),$$ $${y_2}\left( x \right), \ldots ,$$ $${y_n}\left( x \right)$$ are linearly independent. This property of the Wronskian allows to determine whether the solutions of a homogeneous differential equation are linearly independent.

## Fundamental System of Solutions

A set of two linearly independent particular solutions of a linear homogeneous second order differential equation forms its fundamental system of solutions.

If $${y_1}\left( x \right),{y_2}\left( x \right)$$ is a fundamental system of solutions, then the general solution of the second order equation is represented as

$y\left( x \right) = {C_1}{y_1}\left( x \right) + {C_2}{y_2}\left( x \right),$

where $${C_1}, {C_2}$$ are arbitrary constants.

Note that for a given fundamental system of solutions $${y_1}\left( x \right),$$ $${y_2}\left( x \right)$$ we can construct the corresponding homogeneous differential equation. For the case of a second order equation, it is expressed in terms of the determinant:

$\left| {\begin{array}{*{20}{c}} {{y_1}}&{{y_2}}&y\\ {{y'_1}}&{{y'_2}}&y'\\ {{y^{\prime\prime}_1}}&{{y^{\prime\prime}_2}}&y^{\prime\prime} \end{array}} \right| = 0.$

## Liouville's Formula

Thus, as noted above, the general solution of a homogeneous second order differential equation is a linear combination of two linearly independent particular solutions $${y_1}\left( x \right),$$ $${y_2}\left( x \right)$$ of this equation.

Obviously, the particular solutions depend on the coefficients of the differential equation. The Liouville formula establishes a connection between the Wronskian $$W\left( x \right),$$ constructed on the basis of particular solutions $${y_1}\left( x \right),$$ $${y_2}\left( x \right),$$ and the coefficient $${a_1}\left( x \right)$$ in the differential equation.

Let $$W\left( x \right)$$ be the Wronskian of the solutions $${y_1}\left( x \right),$$ $${y_2}\left( x \right)$$ of a linear second order homogeneous differential equation

$y^{\prime\prime} + {a_1}\left( x \right)y' + {a_2}\left( x \right)y = 0,$

in which the functions $${a_1}\left( x \right)$$ and $${a_2}\left( x \right)$$ are continuous on the interval $$\left[ {a,b} \right].$$ Let the point $${x_0}$$ belong to the interval $$\left[ {a,b} \right].$$ Then for all $$x \in \left[ {a,b} \right]$$ the Liouville formula

$W\left( x \right) = W\left( {{x_0}} \right)\exp \left( { - \int\limits_{{x_0}}^x {{a_1}\left( t \right)dt} } \right)$

is valid.

## Practical Methods for Solving Second Order Homogeneous Equations with Variable Coefficients

Unfortunately, there is no general method for finding a particular solution. Usually this is done by guessing.

If a particular solution $${y_1}\left( x \right) \ne 0$$ of the homogeneous linear second order equation is known, the original equation can be converted to a linear first order equation using the substitution $$y = {y_1}\left( x \right)z\left( x \right)$$ and the subsequent replacement $$z'\left( x \right) = u.$$

Another way to reduce the order is based on the Liouville formula. In this case, a particular solution $${y_1}\left( x \right)$$ must also be known. The relevant examples are given below.

## Solved Problems

### Example 1.

Investigate whether the functions ${y_1}\left( x \right) = x + 2, {y_2}\left( x \right) = 2x - 1$ are linearly independent.

Solution.

We form the quotient of two functions:

$\frac{{{y_1}\left( x \right)}}{{{y_2}\left( x \right)}} = \frac{{x + 2}}{{2x - 1}} = \frac{{x - \frac{1}{2} + \frac{5}{2}}}{{2x - 1}} = \frac{{\frac{1}{2}\left( {2x - 1} \right) + \frac{5}{2}}}{{2x - 1}} = \frac{1}{2} + \frac{5}{{2\left( {2x - 1} \right)}} = \frac{1}{2} + \frac{5}{{4x - 2}}.$

It is seen that this ratio is not equal to a constant, but depends on $$x.$$ Hence, these functions are linearly independent.

### Example 2.

Find the Wronskian of the system of functions ${y_1}\left( x \right) = \cos x, {y_2}\left( x \right) = \sin x.$

Solution.

The Wronskian of the system of two functions is calculated by the formula:

${W_{{y_1},{y_2}}}\left( x \right) = \left| {\begin{array}{*{20}{c}} {{y_1}\left( x \right)}&{{y_2}\left( x \right)}\\ {{y'_1}\left( x \right)}&{{y'_2}\left( x \right)} \end{array}} \right|.$

Substituting the given functions and their derivatives, we obtain

${W_{{y_1},{y_2}}}\left( x \right) = \left| {\begin{array}{*{20}{c}} {\cos x}&{\sin x}\\ { - \sin x}&{\cos x} \end{array}} \right| = {\cos ^2}x + {\sin ^2}x = 1.$

It follows from here, that functions $$\sin x$$ and $$\cos x$$ are linearly independent.

See more problems on Page 2.