4.4. LINEAR SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS169

4.4Linear Second Order Homogeneous Differential Equations

Definition 4.4.1 A linear second order di erential equation in the variable y is an equation of the form

y00 + p(x)y0 + q(x)y = r(x).

If r(x) = 0, we say that the equation is homogeneous; otherwise it is called non-homogeneous. If p(x) and q(x) are constants, we say that the equation has constant coe cients.

Definition 4.4.2 If f and g are di erentiable functions, then the Wronskian of f and g is denoted W (f, g) and defined by

W (f, g) = f(x)g0(x) − f0(x)g(x).

Example 4.4.1 Compute the following Wronskians:

Part (i) W (sin mx, cos mx) = sin(mx)dxd (cos(mx)) − dxd (sin(mx)) cos(mx) = −m sin2(mx) − m cos2(mx)

= −m(sin2(mx) + cos2(mx)) = −m

Part (iii) W (epx sin qx, epx cos qx)

=epx sin qx(pepx cos qx − qepx sin qx)

−epx cos qx(pepx sin qx + qepx cos qx)

=−qe2px(sin2 qx + cos2 qx)

=−qe2px.

Part (iii) W (xn, xm) = xn · mxm−1 − xm · nxn−1

= (m − n)xn+m−1.

Part (iv) W (x sin mx, x cos mx) = (x sin mx)(cos mx − mx sin mx) −(x cos mx)(sin mx + mx cos mx)

=−mx2(sin2 mx + cos2 mx)

=−mx2.

Definition 4.4.3 Two di erentiable functions f and g are said to be linearly independent if their Wronskian, W (f(x), g(x)), is not zero for all x in the domains of both f and g.

Example 4.4.2 Which pairs of functions in Example 8 are linearly independent?

(i) In Part (i), W (sin mx, cos mx) = −m 6= 0 unless m = 0. Therefore, sin mx and cos mx are linearly independent if m 6= 0.

(ii) In Part (ii),

W (epx sin qx, epx cos qx) = −qe2px 6≡if q 6= 0.

Therefore, epx sin(qx) and epx cos qx are linearly independent if q 6= 0.

4.4.LINEAR SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS171

(iii)In Part (iii), W (xn, xm) = (m − n)xn+m−1 6≡0 if m 6= n. Therefore, if m and n are not equal, then xn and xm are linearly independent.

(iv)In Part (iv),

W (x sin mx, x cos mx) = −mx2 6≡0

if m 6= 0. Therefore, x sin mx and x cos mx are linearly independent if m 6= 0.

Theorem 4.4.1 Consider the linear homogeneous second order di erential equation

(i)If y1(x) and y2(x) are any two solutions of (1), then every linear combination y(x), with constants A and B,

y(x) = Ay1(x) + By2(x)

is also a solution of (1).

(ii)If y1(x) and y2(x) are any two linearly independent solutions of (1), then every solution y(x) of (1) has the form

y(x) = Ay1(x) + By2(x)

for some constants A and B.

Proof.

Part (i) Suppose that y1 and y2 are solutions of (1), A and B are any constants. Then

(Ay1 + By2)00 + p(Ay1 + By2)0 + q(Ay1 + By2)

=Ay100 + By200 + Apy10 + ABy20 + Aqy1 + BqBy2

=A(y100 + py10 + qy1) + B(y200 + py20 + qy2)

=A(0) + B(0) (Because y1 and y2 are solutions of (1))

=0.

Hence, y = Ay1 + By2 are solutions of (1) whenever y1 and y2 are solutions of (1).

We solve for A and B from equations (2) and (3) to get

Since y1 and y2 are linearly independent, W (y1, y2) 6= 0, and hence, A and B are uniquely determined.

Remark 15 It turns out that the Wronskian of two solutions of (1) is either identically zero or never zero for any value of x.

Theorem 4.4.2 Let y1 and y2 be any two solutions of the homogeneous equation

Let

W (x) = W (y1, y2) = y1(x)y20 (x) − y10 (x)y2(x).

Then

W 0(x) = −pW (x)

R

W (x) = ce− p(x)dx

for some constant c. If c = 0, then W (x) = 0 for every x. If c 6= 0, then

4.4. LINEAR SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS173

R Z R

W (x) = e− pdx 0 dx + c = ce− pdx.

If c = 0, W (x) ≡ 0; otherwise W (x) is never zero.

Theorem 4.4.3 (Homogeneous Second Order) Consider the linear second order homogeneous di erential equation with constant coe cients:

ing three cases arise:

Case 1. The discriminant b2 − 4ac > 0. Then m1 and m2 are real and distinct. The two linearly independent solutions of (1) are em1x and

and its general solution has the form

y(x) = Aem1x + Bem2x.

Case 2. The discriminant b2 − 4ac = 0. Then m1 = m2 = m, and only one real solution exists for equation (2). The roots are repeated. In this case, emx and xemx are two linearly independent solutions of (1) and the general solution of (1) has the form

y(x) = Aemx + Bxemx = emx(A + Bx).

Case 1. For Case 1, em1x and em2x are solutions of (1). We show that these are linearly independent by showing that their Wronskian is not zero.

W (em1x, em2x) = em1x · m2em2x − m1em1x · em2x

ay00 + by0 + cy = a(2m + m2x)emx + b(1 + mx)emx + xemx = (b + 2am)emx + (am2 + bm + c)xemx

= (b + 2a(−b/2a))emx

= 0.

176 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

Remark 16 Use Integral Tables or computer algebra in evaluating the indefinite integrals as needed.

Example 4.4.3 Solve the di erential equations for y(t).

(i)y00 − 5y0 + 14y = 0

(ii)y00 − 6y0 + 9y = 0

(iii)y00 − 4y0 + 5y = 0

We let y = emt. We then solve for m and determine the solution. We observe that y0 = memt, y00 = m2emt.

Part (i) By substituting y = emt in the equation we get

m2emt − 5memt + 14emt = emt(m2 − 5m + 14) = 0 → m2 − 5m + 14 = 0 = (m − 7)(m + 2) → m = 7, −2.

Therefore,

y(t) = Ae−2t + Be7t.

Part (ii) Again, by substituting y = emt, we get

m2emt − 6memt + 9emt = 0

m2 − 6m + 9 = (m − 3)2 = 0 m = 3, 3.

The solution is

y(t) = Ae3t + Bte3t.

Part (iii) By substituting y = emt, we get

m2emt − 4memt + 5emt = emt(m2 − 4m + 5) = 0 →

m2 − 4m + 5 = 0

√

m = 4 ± 16 − 20 = 2 ± 1i. 2

The general solution is

y(t) = e2t(A cos t + B sin t).

4.4. LINEAR SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS177

Example 4.4.4 Solve the di erential equation for y(t) where y(t) satisfies the conditions

y00(t) − 2y0(t) − 15y(t) = 0; y(0) = 1, y0(0) = −1.

We assume that y(t) = emt. By substitution we get the characteristic equa-

tion:

m2 − 2m − 15 = 0, m = 5, −3.

The general solution is

y(t) = Ae−3t + Be5t.

We now impose the additional conditions y(0) = 1, y0(0) = −1.

y(t) = Ae−3t + Be5t y0(t) = −3Ae−3t + 5Be5t y(0) = A + B = 1

y0(0) = −3A + 5B = −1

On solving these two equations simultaneously, we get

A = 43, B = 14 .

y(t) = 34 e−3t + 14 e5t.

Exercises 4.4 Solve for y(t) from each of the following:

1.y00 − y0 − 20y = 0

2.y00 − 8y0 + 16y = 0

3.y00 + 9y0 + 20y = 0

4.y00 + 4y0 + 4 = 0

5.y00 − 8y0 + 12y = 0

6.y00 − 6y0 + 10y = 0

7.y0 − y0 − 6y = 0, y(0) = 10, y0(0) = 15

8.y00 − 4y0 + 4y − 0, y(0) = 4, y0(0) = 8

9.y00 + 8y0 + 12y = 0, y(0) = 1, y0(0) = 3

10.y00 + 6y0 + 10y = 0, y(0) = 5, y0(0) = 7

11.y00 − 4y = 0, y(0) = 1, y0(0) = −1

12.y00 − 9y = 0, y(0) = −1, y0(0) = 1

13.y00 + 9y = 0, y(0) = 2, y0(0) = 3

14.y00 + 4y = 0, y(0) = −1, y0(0) = 2

15.y00 − 3y0 + 2y = 0, y(0) = 2, y0(0) = −2

16.y00 − y0 − 6y = 0, y(0) = 6, y0(0) = 5

17.y00 + 4y0 + 4y = 0, y(0) = 1, y0(0) = 4

18.y00 − 6y0 + 9y = 0, y(0) = 1, y0(0) = −1

19.y00 + 6y0 + 13y = 0, y(0) = 1, y0(0) = 2

20.y00 − 3y0 + 2y = 0

21.y00 + 3y0 + 2y = 0

22.y00 + m2y = 0

23.y00 − m2y = 0

24.y00 + 2my0 + m2y2 = 0

25.y00 + 2my0 + (m2 + 1)y = 0

26.y00 − 2my0 + (m2 + 1)y = 0

27.y00 + 2my0 + (m2 − 1)y = 0

28.y00 − 2my0 + (m2 − 1)y = 0

29.9y00 − 12y0 + 4y = 0

30.4y00 + 4y0 + y = 0