Engenharia

SECOND-ORDER DIFFERENTIAL EQUATIONS

17.1
Second-Order Linear Equations
In this section, we will learn how to solve:
Homogeneous linear equations for various cases and for initial- and boundary-value problems.

SECOND-ORDER LINEAR EQNS.

Equation 1

A second-order linear differential equation has the form 2

d y dy P  x 2  Q  x  R  x y  G  x dx dx

where P, Q, R, and G are continuous functions.

In this section, we study the case where G(x) = 0, for all x, in Equation 1.
– Such equations are called homogeneous linear equations.

HOMOGENEOUS LINEAR EQNS.

Equation 2

Thus, the form of a second-order linear homoge-neous differential equation is:

d x dy P  x 2  Q  x  R  x y  0 dy dx
If G(x) ≠ 0 for some x, Equation 1 is nonhomogeneous and is discussed in Section 17.2.

2

Two basic facts enable us to solve homogeneous linear equations.

The first says that, if we know two solutions y1 and y2 of such an equation, then the linear combination y = c1y1 + c2y2 is also a solution.

SOLVING HOMOGENEOUS EQNS.

Theorem 3

If y1(x) and y2(x) are both solutions of the linear homogeneous equation 2 and c1 and c2 are any constants, then the function

y(x) = c1y1(x) + c2y2(x) is also a solution of Equation 2. The other fact we need is given by the following theorem, which is proved in more advanced courses.
– It says that the general solution is a linear combination of two linearly independent solutions y1 and y2.

SOLVING HOMOGENEOUS EQNS.

This means that neither y1 nor y2 is a constant multiple of the other.
–For instance, the functions f(x) = x2 and g(x) = 5x2 are L D, but f(x) = ex and g(x) = xex are

L I.

Theorem 4 If y1 and y2 are linearly independent solutions of Equation 2, and P(x) is never 0, then the general solution is given by:

y(x) = c1y1(x) + c2y2(x) where c1 and c2 are arbitrary constants.

SOLVING HOMOGENEOUS EQNS.

Theorem 4 is very useful because it says that, if we know two