Second Order Linear Homogeneous Differential Equations with Variable Coefficients

Trigonometry

Trigonometry Logo

Second Order Linear Homogeneous Differential Equations with Variable Coefficients

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

y+a1(x)y+a2(x)y=0,

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

Linear Independence of Functions. Wronskian

The functions y1(x),y2(x),,yn(x) are called linearly dependent on the interval [a,b], if there are constants α1,α2,,αn, not all zero, such that for all values of x from this interval, the identity

α1y1(x)+α2y2(x)++αnyn(x)0

holds. If this identity is satisfied only when α1=α2= =αn=0, then these functions y1(x),y2(x),, yn(x) are called linearly independent on the interval [a,b].

For the case of two functions, the linear independence criterion can be written in a simpler form: The functions y1(x), y2(x) are linearly independent on the interval [a,b], if their quotient in this segment is not identically equal to a constant:

y1(x)y2(x)const.

Otherwise, when y1(x)y2(x)const, these functions are linearly dependent.

Let n functions y1(x), y2(x),, yn(x) have derivatives of (n1) order. The determinant

W(x)=Wy1,y2,,yn(x)=|y1y2yny1y2yny1(n1)y2(n1)yn(n1)|

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

Wronskian Test.

If the system of functions y1(x), y2(x),, yn(x) is linearly dependent on the interval [a,b], 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 [a,b], then the functions y1(x), y2(x),, yn(x) 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 y1(x),y2(x) is a fundamental system of solutions, then the general solution of the second order equation is represented as

y(x)=C1y1(x)+C2y2(x),

where C1,C2 are arbitrary constants.

Note that for a given fundamental system of solutions y1(x), y2(x) we can construct the corresponding homogeneous differential equation. For the case of a second order equation, it is expressed in terms of the determinant:

|y1y2yy1y2yy1y2y|=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 y1(x), y2(x) 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(x), constructed on the basis of particular solutions y1(x), y2(x), and the coefficient a1(x) in the differential equation.

Let W(x) be the Wronskian of the solutions y1(x), y2(x) of a linear second order homogeneous differential equation

y+a1(x)y+a2(x)y=0,

in which the functions a1(x) and a2(x) are continuous on the interval [a,b]. Let the point x0 belong to the interval [a,b]. Then for all x[a,b] the Liouville formula

W(x)=W(x0)exp(x0xa1(t)dt)

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 y1(x)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=y1(x)z(x) and the subsequent replacement z(x)=u.

Another way to reduce the order is based on the Liouville formula. In this case, a particular solution y1(x) must also be known. The relevant examples are given below.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Investigate whether the functions y1(x)=x+2,y2(x)=2x1 are linearly independent.

Example 2

Find the Wronskian of the system of functions y1(x)=cosx,y2(x)=sinx.

Example 1.

Investigate whether the functions y1(x)=x+2,y2(x)=2x1 are linearly independent.

Solution.

We form the quotient of two functions:

y1(x)y2(x)=x+22x1=x12+522x1=12(2x1)+522x1=12+52(2x1)=12+54x2.

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 y1(x)=cosx,y2(x)=sinx.

Solution.

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

Wy1,y2(x)=|y1(x)y2(x)y1(x)y2(x)|.

Substituting the given functions and their derivatives, we obtain

Wy1,y2(x)=|cosxsinxsinxcosx|=cos2x+sin2x=1.

It follows from here, that functions sinx and cosx are linearly independent.