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17.1 Second-Order Linear

Equations

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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.

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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.

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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.

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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) = ex and g(x) = xex are linearly independent.

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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.

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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.

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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 = erx (where r is a constant) has the property that its derivative is a constant multiple of itself: y′ = rerx. Furthermore, y″ = r2erx. If we substitute these expressions into Equation 5, we see that y = erx is a solution if

ar2erx + brerx + cerx = 0 or (ar2 + br + c)erx = 0

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Second-Order Linear Equations

But erx is never 0. Thus y = erx 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.

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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.

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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.

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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

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Example 1 – Solution

We could verify that this is indeed a solution by

differentiating and substituting into the differential equation.

cont’d

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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 = erx is one solution of Equation 5. We now verify that y2 = xerx is also a solution:

= a(2rerx + r2xerx) + b(erx + rxerx) + cxerx

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Second-Order Linear Equations

= 0(erx) + 0(xerx) = 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 = erx and y2 = xerx are linearly independent

solutions, Theorem 4 provides us with the general solution.

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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

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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

eiθ = cos θ + i sin θ

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Second-Order Linear Equations

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

= C1eαx(cos βx + i sin βx) + C2eαx(cos βx – i sin βx)

= eαx[(C1 + C2) cos βx + i(C1 – C2) sin βx]

= eαx(c1 cos βx + c2 sin βx) where c = C + C , c = i(C – C ).

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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.

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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

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Example 4 – Solution

By (11), the general solution of the differential equation is y = e3x(c1 cos 2x + c2 sin 2x)

cont’d

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Initial-Value and

Boundary-Value Problems

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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.

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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) = c1e2x + c2e–3x Differentiating this solution, we get

y′(x) = 2c e2x – 3c e–3x

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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

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Example 5 – Solution

Thus the required solution of the initial-value problem is y = e2x + e–3x

cont’d

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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.

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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

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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

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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

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