- •Functions
- •The Concept of a Function
- •Trigonometric Functions
- •Inverse Trigonometric Functions
- •Logarithmic, Exponential and Hyperbolic Functions
- •Limits and Continuity
- •Introductory Examples
- •Continuity Examples
- •Linear Function Approximations
- •Limits and Sequences
- •Properties of Continuous Functions
- •The Derivative
- •The Chain Rule
- •Higher Order Derivatives
- •Mathematical Applications
- •Antidifferentiation
- •Linear Second Order Homogeneous Differential Equations
- •Linear Non-Homogeneous Second Order Differential Equations
- •Area Approximation
- •Integration by Substitution
- •Integration by Parts
- •Logarithmic, Exponential and Hyperbolic Functions
- •The Riemann Integral
- •Volumes of Revolution
- •Arc Length and Surface Area
- •Techniques of Integration
- •Integration by formulae
- •Integration by Substitution
- •Integration by Parts
- •Trigonometric Integrals
- •Trigonometric Substitutions
- •Integration by Partial Fractions
- •Fractional Power Substitutions
- •Numerical Integration
- •Integrals over Unbounded Intervals
- •Discontinuities at End Points
- •Improper Integrals
- •Sequences
- •Monotone Sequences
- •Infinite Series
- •Series with Positive Terms
- •Alternating Series
- •Power Series
- •Taylor Polynomials and Series
- •Applications
- •Parabola
- •Ellipse
- •Hyperbola
- •Polar Coordinates
- •Graphs in Polar Coordinates
- •Areas in Polar Coordinates
- •Parametric Equations
4.4. LINEAR SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS169
15. |
y0 |
= x |
− |
y, y(0) = 5 |
16. |
y0 |
+ 3x2y = 0, y(0) = 6 |
||||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
||
17. |
xy0(x) + 3y(x) = 2x5, y(2) = 1 |
18. |
xy0 + y = 3x2, y(1) = 4 |
||||||||||||
19. |
y0 |
+ y = ex, y(0) = 100 |
20. |
y0 |
= −6xy, y(0) = 9 |
||||||||||
21. |
y0 |
= (sin x)y, y(0) = 5 |
22. |
y0 |
= xy3, y(0) = 2 |
||||||||||
|
|
1 + |
√ |
|
|
|
|
|
|
|
|
||||
|
y0 |
x |
|
y0 − 2y = 1, y(1) = 3 |
|||||||||||
23. |
= |
|
|
|
|
|
|
, y(0) = 10 |
24. |
||||||
1 + |
√ |
|
|
||||||||||||
y |
|||||||||||||||
25. |
y0 |
= ry − c, y(0) = A |
26. |
y0 |
− 3y = 2 sin x, y(0) = 12 |
||||||||||
27. |
y0 |
− 2y = 4e2x, y(0) = 4 |
28. |
y0 |
− 3x2y = ex3 , y(0) = 7 |
||||||||||
|
y0 |
1 |
|
|
|
|
|
|
|
|
|
y0 |
− 3y = e2x, y(0) = 1 |
||
29. |
− |
|
y = sin x, y(1) = 3 |
30. |
|||||||||||
2x |
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:
170 |
CHAPTER 4. |
APPLICATIONS OF DIFFERENTIATION |
(i) |
W (sin(mx), cos(mx)) |
(ii) W (epx sin(qx), epx cos(qx)) |
(iii) |
W (xn, xm) |
(iv) W (x sin(mx), x cos(mx)) |
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
y00 + p(x)y0 + q(x)y = 0. |
(1) |
(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).
172 |
CHAPTER 4. APPLICATIONS OF DIFFERENTIATION |
||
Part (ii) Let y be any solution of (1) and suppose that |
|
||
|
y = Ay1 |
+ By2 |
(2) |
|
y0 = Ay10 |
+ By20 |
(3) |
We solve for A and B from equations (2) and (3) to get
A = |
yy20 − y2y0 |
= |
W (y, y2) |
|
||
|
y1y20 |
− y2y10 |
|
W (y1, y2) |
||
B = |
y1y0 − y10 y |
= |
W (y1, y) |
. |
||
|
|
|||||
|
y1y20 |
− y2y10 |
|
(y1, y2) |
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
y00 + py0 + qy = 0. |
(1) |
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
W (x) 6= 0 for every x. |
|
|
|
Proof. Since y1 and y2 are solutions of (1), |
|
||
y100 |
= −py10 |
− qy1 |
(2) |
y200 |
= −py20 |
− qy2 |
(3) |
4.4. LINEAR SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS173
Then, |
|
W 0(x) = (y1y20 − y10 y2)0 |
|
= y10 y20 + y1y200 − y100y2 − y10 y20 |
|
= y1y200 − y2y100 |
|
= y1(−py20 − qy2) − y2(−py10 − qy1) |
(from (2) and (3)) |
= −p[y1y20 − y2y10 ] |
|
= −pW (x). |
|
Thus, |
|
W 0(x) + pW (x) = 0. |
|
By Theorem 4.3.1 |
|
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:
|
|
|
ay00 + by0 + cy = 0, a 6= 0. |
(1) |
|||||
(i) If y = emx is a solution of (1), then |
|
|
|
|
|||||
|
|
|
|
am2 + bm + c = 0. |
|
|
(2) |
||
Equation (2) is called the characteristic equation of (1). |
|||||||||
|
√ |
|
|
|
√ |
|
|
|
|
2 |
|
b |
2 |
|
|
||||
(ii) Let m1 = |
−b − b |
− 4ac |
and m2 = |
−b + |
|
− 4ac |
. Then the follow- |
||
|
|
||||||||
2a |
|
2a |
|
||||||
|
|
|
|
|
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.
174 |
CHAPTER 4. APPLICATIONS OF DIFFERENTIATION |
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 3 b2 − 4ac < 0. Then m1 = p − iq, and m2 = p + iq where p = |
||||||||||
b/ a, and q = |
√ |
4ac |
− |
b2 |
/2a. In this case, the functions epx sin qx |
|||||
− 2 px |
|
|
|
|
|
|
|
|
|
|
and e |
cos qx are two linearly independent solutions of (1) and the |
|||||||||
most general solution of (1) has the form |
||||||||||
|
|
|
y(x) = epx(A sin qx + B sin qx). |
|||||||
Proof. Let y = emx. Then y0 |
= memx, y00 = m2emx and |
|||||||||
ay00 + by0 + cy = (am2 + bm + c)emx = 0, a 6= 0 ↔ |
||||||||||
am2 + bm + c = 0, a 6= 0 ↔ |
||||||||||
|
|
√ |
|
|
|
|||||
m = |
−b |
b2 |
|
− 4ac |
. |
|||||
2a ± |
|
|
||||||||
|
|
2a |
|
|
|
|||||
This proves Part (i). |
|
|
|
|
|
|
|
|
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
|
= (m2 − m1)e(m1+m2)x. |
|
|
Since m1 6= m2, W (em1x, em2x) 6= 0. |
|
|
|
We already know that emx = 0, and m = |
− |
b/2a. Let us try |
|
Case 2. mx |
6 |
|
|
y = xe |
. Then |
|
|
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.
4.4. LINEAR SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS175
Therefore, emx and xemx are both solutions. We only need to show that they are linearly independent.
W (emx, xemx) = emx(emx + mxemx) − memx(xemx)
=e2mx + mxe2mx − mxe2mx
=e2mx
6= 0.
Hence, emx and xemx are linearly independent and the general solution of (1) has the form
y(x) = Aemx + Bxemx = emx(A + Bx).
Case 3. In Example 8, we showed that
W (epx sin qx, epx cos qx) = −qe2px 6= 0
since q 6= 0.
We only need to show that epx sin qx and epx cos qx are solutions of (1). Let y1 = epx sin qx and y2 = epx cos qx. Then,
y10 |
= pepx sin qx + qepx cos qx |
|
|
|
|
|
|
|
|
|
|
|
|
||||
y100 |
= p2epx sin qx + pqepx cos qx + pqepx cos qx − q2 sin qx |
||||||||||||||||
ay100 + by10 + cy1 = aepx(p2 sin qx + 2pq cos qx − q2 sin qx) |
|
|
|
||||||||||||||
|
+ bepx(p sin qx + q cos qx) + cepx sin qx |
|
|
|
|||||||||||||
|
= epx sin qx[a(p2 − q2) + (bp + c)] + epx cos qx[2apq + bq] |
||||||||||||||||
|
= epx sin(qx) a |
|
b2 |
|
4ac − b2 |
|
|
+ b |
−b |
+ c |
|||||||
|
|
|
|
|
|
2a |
|||||||||||
|
4a2 − |
|
4a2 |
|
|
|
|
||||||||||
|
|
|
|
2a |
|
√ |
|
− |
|
|
|
||||||
|
+ epx cos qx |
|
−b |
|
|
|
|
|
|
|
|
||||||
|
2a |
|
|
+ b |
|
4ac |
|
b2 |
|
||||||||
|
|
b2 |
|
ac + b2 + 2ac |
|
|
|
|
|||||||||
|
= epx sin(qx) |
|
− 2 |
2a |
|
|
|
|
|
|
|
|
|
||||
|
|
|
|
|
|
|
|
|
|
|
+ epx cos(qx)[0] = 0.
Therefore y1 = epx sin(qx) is a solution of (1). Similarly, we can show that y2 = epx cos qx is a solution of (1). We leave this as an exercise.
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
178 |
CHAPTER 4. APPLICATIONS OF DIFFERENTIATION |
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