17.1

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 y₁ and y₂ of such an equation, then the linear combination

y = c₁y₁ + c₂y₂ 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 y₁ and y₂.

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

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″ = r²e^rx. If we substitute these expressions into Equation 5, we see that y = e^rx is a solution if

ar²e^rx + bre^rx + ce^rx = 0 or (ar² + br + c)e^rx = 0

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 r², y′ by r, and y by 1.

Second-Order Linear Equations

Sometimes the roots r₁ and r₂ 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 b² – 4ac.

Second-Order Linear Equations

Case I: b² – 4ac > 0

In this case the roots r₁ and r₂ of the auxiliary equation are real and distinct, so y₁ = and y₂ = 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

r² + 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: b² – 4ac = 0

In this case r₁ = r₂; that is, the roots of the auxiliary equation are real and equal. Let’s denote by r the common value of r₁ and r₂. Then, from Equations 7, we have

so 2ar + b = 0

We know that y₁ = e^rx is one solution of Equation 5. We now verify that y₂ = xe^rx is also a solution:

= a(2re^rx + r²xe^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, ar² + br + c = 0 because r is a root of the auxiliary equation.

Since y₁ = e^rx and y₂ = 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 4r² + 12r + 9 = 0 can be factored as (2r + 3)² = 0

so the only root is r = By (10) the general solution is y = c₁e^–3x/2 + c₂xe^–3x/2

Second-Order Linear Equations

Case III: b² – 4ac < 0

In this case the roots r₁ and r₂ of the auxiliary equation are complex numbers. We can write

r₁ = α + iβ r₂ = α – iβ

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

e^iθ = cos θ + i sin θ

Second-Order Linear Equations

We write the solution of the differential equation as y = C₁ + C₂ = C₁e^{(α +iβ)x} + C₂e^{(α – iβ)x}

= C₁e^αx(cos βx + i sin βx) + C₂e^αx(cos βx – i sin βx)

= e^αx[(C₁ + C₂) cos βx + i(C₁ – C₂) sin βx]

= e^αx(c₁ cos βx + c₂ sin βx) 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 c₁ and c₂ are real.

We summarize the discussion as follows.

Example 4

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

Solution:

The auxiliary equation is r² – 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(c₁ cos 2x + c₂ 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(x₀) = y0 y′(x₀) = y₁

where y₀ and y₁ 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) = c₁e^2x + c₂e^–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) = c₁ + c₂ = 1

y′(0) = 2c₁ – 3c₂ = 0

From (13), we have c₂ = c₁ and so (12) gives c₁ + c₁ = 1 c₁ = c₂ =

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(x₀) = y0 y(x₁) = y₁

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

r² + 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) = c₁ = 1

y(1) = c₁e^–1 + c₂e^–1 = 3

The first condition gives c₁ = 1, so the second condition becomes

e^–1 + c₂e^–1 = 3

cont’d

Example 7 – Solution

Solving this equation for c₂ by first multiplying through by e, we get

1 + c₂ = 3e so c₂ = 3e – 1

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

cont’d