**17.1** Second-Order Linear

### Equations

### Second-Order Linear Equations

**A second-order linear differential equation has the form**

*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. **

### Second-Order Linear Equations

Thus the form of a second-order linear homogeneous differential equation is

*If G(x) ≠ 0 for some x, Equation 1 is nonhomogeneous.*

### Second-Order Linear Equations

Two basic facts enable us to solve homogeneous linear equations.

*The first of these says that if we know two solutions y*_{1 }and
*y*_{2 }**of such an equation, then the linear combination**

*y = c*_{1}*y*_{1 }*+ c*_{2}*y*_{2 }is also a solution.

### Second-Order Linear Equations

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 y**_{1 }*and y*_{2}.

*This means that neither y*_{1 }*nor y*_{2} is a constant multiple of
the other.

*For instance, the functions f(x) = x*^{2} *and g(x) = 5x*^{2} are

*linearly dependent, but f(x) = e*^{x}*and g(x) = xe** ^{x}* are linearly
independent.

### Second-Order Linear Equations

Theorem 4 is very useful because it says that if we know
*two particular linearly independent solutions, then we know *
*every solution.*

In general, it’s not easy to discover particular solutions to a second-order linear equation.

### Second-Order Linear Equations

But it is always possible to do so if the coefficient functions
*P, Q, and R are constant functions, that is, if the differential *
equation has the form

*where a, b, and c are constants and a* ≠ 0.

It’s not hard to think of some likely candidates for particular solutions of Equation 5 if we state the equation verbally.

### Second-Order Linear Equations

*We are looking for a function y such that a constant times*
*its second derivative y″ plus another constant times y′ plus *
*a third constant times y is equal to 0.*

*We know that the exponential function y = e*^{rx}*(where r is a *
constant) has the property that its derivative is a constant
*multiple of itself: y′ = re*^{rx}*. Furthermore, y″ = r*^{2}*e** ^{rx}*. If we
substitute these expressions into Equation 5, we see that

*y = e*

*is a solution if*

^{rx}*ar*^{2}*e*^{rx}*+ bre*^{rx}*+ ce** ^{rx}* = 0
or

*(ar*

^{2}

*+ br + c)e*

*= 0*

^{rx}### Second-Order Linear Equations

*But e*^{rx}*is never 0. Thus y = e** ^{rx}* is a solution of Equation 5 if

*r is a root of the equation*

**Equation 6 is called the auxiliary equation (or **

**characteristic equation) of the differential equation**
*ay″ + by′ + cy = 0.*

Notice that it is an algebraic equation that is obtained from
*the differential equation by replacing y″ by r*^{2}*, y′ by r, and y *
by 1.

### Second-Order Linear Equations

*Sometimes the roots r*_{1 }*and r*_{2 }of the auxiliary equation can
be found by factoring. In other cases they are found by

using the quadratic formula:

We distinguish three cases according to the sign of the
*discriminant b*^{2} *– 4ac.*

### Second-Order Linear Equations

**Case I: b**^{2}**– 4ac > 0**

*In this case the roots r*_{1 }*and r*_{2 }of the auxiliary equation are
*real and distinct, so y*_{1 }*= and y*_{2 }= are two linearly
independent solutions of Equation 5. (Note that is not a
constant multiple of .)

Therefore, by Theorem 4, we have the following fact.

### Example 1

*Solve the equation y″ + y′ – 6y = 0.*

Solution:

The auxiliary equation is

*r*^{2} *+ r – 6 = (r – 2)(r + 3) = 0*
*whose roots are r = 2, –3.*

Therefore, by (8), the general solution of the given differential equation is

*Example 1 – Solution*

We could verify that this is indeed a solution by

differentiating and substituting into the differential equation.

cont’d

### Second-Order Linear Equations

**Case II: b**^{2}**– 4ac = 0**

*In this case r*_{1 }*= r*_{2}; that is, the roots of the auxiliary equation
*are real and equal. Let’s denote by r the common value of*
*r*_{1 }*and r*_{2}. Then, from Equations 7, we have

*so 2ar + b = 0 *

*We know that y*_{1 }*= e** ^{rx }*is one solution of Equation 5. We

*now verify that y*

_{2 }

*= xe*

*is also a solution:*

^{rx }*= a(2re*^{rx}*+ r*^{2}*xe*^{rx}*) + b(e*^{rx}*+ rxe*^{rx}*) + cxe*^{rx}

### Second-Order Linear Equations

*= 0(e*^{rx}*) + 0(xe** ^{rx}*) = 0

*In the first term, 2ar + b = 0 by Equations 9; in the second *
*term, ar*^{2}*+ br + c = 0 because r is a root of the auxiliary *
equation.

*Since y*_{1 }*= e*^{rx }*and y*_{2 }*= xe** ^{rx }*are linearly independent

solutions, Theorem 4 provides us with the general solution.

### Example 3

*Solve the equation 4y″ + 12y′ + 9y = 0.*

Solution:

*The auxiliary equation 4r*^{2} *+ 12r + 9 = 0 can be factored as*
*(2r + 3)*^{2} = 0

*so the only root is r = By (10) the general solution is*
*y = c*_{1}*e*^{–3x/2}*+ c*_{2}*xe*^{–3x/2}

### Second-Order Linear Equations

**Case III: b**^{2}**– 4ac < 0**

*In this case the roots r*_{1 }*and r*_{2 }of the auxiliary equation are
complex numbers. We can write

*r*_{1} = α *+ i*β *r*_{2} = α *– i*β

where α and β are real numbers. [In fact, α *= –b/(2a),*
β = ] Then, using Euler’s equation

*e*^{i}^{θ} = cos θ *+ i sin *θ

### Second-Order Linear Equations

We write the solution of the differential equation as
*y = C*_{1 } *+ C*_{2 } *= C*_{1}*e*^{(}^{α} ^{+i}^{β}^{)x}*+ C*_{2}*e*^{(}^{α} ^{– i}^{β}^{)x}

*= C*_{1}*e*^{α}* ^{x}*(cos β

*x + i sin*β

*x) + C*

_{2}

*e*

^{α}

*(cos β*

^{x}*x – i sin*β

*x)*

*= e*^{α}^{x}*[(C*_{1 }*+ C*_{2}) cos β*x + i(C*_{1} *– C*_{2}) sin β*x]*

*= e*^{α}^{x}*(c*_{1 }cos β*x + c*_{2 }sin β*x)*
*where c* *= C* *+ C* *, c* *= i(C* *– C* ).

### Second-Order Linear Equations

This gives all solutions (real or complex) of the differential
*equation. The solutions are real when the constants c*_{1 }and
*c*_{2 }are real.

We summarize the discussion as follows.

### Example 4

*Solve the equation y″ – 6y′ + 13y = 0.*

Solution:

*The auxiliary equation is r*^{2} *– 6r + 13 = 0. By the quadratic *
formula, the roots are

*Example 4 – Solution*

By (11), the general solution of the differential equation is
*y = e*^{3x}*(c*_{1} *cos 2x + c*_{2} *sin 2x)*

cont’d

### Initial-Value and

### Boundary-Value Problems

### Initial-Value and Boundary-Value Problems

**An initial-value problem for the second-order Equation 1 **
*or 2 consists of finding a solution y of the differential *

equation that also satisfies initial conditions of the form
*y(x*_{0}*) = y*0 *y′(x*_{0}*) = y*_{1}

*where y*_{0 }*and y*_{1 }are given constants.

*If P, Q, R, and G are continuous on an interval and P(x) ≠ 0 *
there, then a theorem found in more advanced books

guarantees the existence and uniqueness of a solution to this initial-value problem. Examples 5 illustrate the

technique for solving such a problem.

### Example 5

Solve the initial-value problem

*y″ + y′ – 6y = 0 y(0) = 1 y′(0) = 0*
Solution:

From Example 1 we know that the general solution of the differential equation is

*y(x) = c*_{1}*e*^{2x}*+ c*_{2}*e** ^{–3x}*
Differentiating this solution, we get

*y′(x) = 2c* *e*^{2x}*– 3c* *e*^{–3x}

*Example 5 – Solution*

To satisfy the initial conditions we require that
*y(0) = c*_{1} *+ c*_{2} = 1

*y′(0) = 2c*_{1} *– 3c*_{2} = 0

*From (13), we have c*_{2 }*= c*_{1 }and so (12) gives
*c*_{1 }*+ c*_{1 }*= 1 c*_{1 }*= c*_{2 }=

cont’d

*Example 5 – Solution*

Thus the required solution of the initial-value problem is
*y = e*^{2x}*+ e*^{–3x}

cont’d

### Initial-Value and Boundary-Value Problems

**A boundary-value problem for Equation 1 or 2 consists of **
*finding a solution y of the differential equation that also *

satisfies boundary conditions of the form

*y(x*_{0}*) = y*0 *y(x*_{1}*) = y*_{1}

In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution.

The method is illustrated in Example 7.

### Example 7

Solve the boundary-value problem

*y″ + 2y′ + y = 0 y(0) = 1 y(1) = 3*
Solution:

The auxiliary equation is

*r*^{2} *+ 2r + 1 = 0 or (r + 1)*^{2} = 0
*whose only root is r = –1.*

Therefore the general solution is

*Example 7 – Solution*

The boundary conditions are satisfied if
*y(0) = c*_{1} = 1

*y(1) = c*_{1}*e*^{–1} *+ c*_{2}*e*^{–1} = 3

*The first condition gives c*_{1} = 1, so the second condition
becomes

*e*^{–1} *+ c*_{2}*e*^{–1 }= 3

cont’d

*Example 7 – Solution*

*Solving this equation for c*_{2 }*by first multiplying through by e, *
we get

*1 + c*_{2 }*= 3e so c*_{2 }*= 3e – 1*

Thus the solution of the boundary-value problem is
*y = e*^{–x}*+ (3e – 1)xe*^{–x}

cont’d