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Differential equations

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1 Differential equations
1.1 Classification of differential equations
◦ A differential equation is an equation that contains an unknown function and one or more
derivatives of the function. The solution set of a given differential equation is a set of functions.
Remark: Regarding the domain and range of the unknown function; in general the independent
variable of the function may be real, complex, a vector or a matrix while the dependent variable
too may be real, complex, a vector or matrix valued. In this class we mainly focus on real valued
functions of a single real variable.
d2 y
dy
or y 0 (x) for the first derivative, 2 or y 00 (x) for the secRemark: About the notation; we write
dx
dx
dn y
(n)
ond derivative · · · , n or y (x) for the nth derivative. Sometimes instead of y 0 (x), y 00 (x), · · · , y (n) (x)
dx
we may simply write y 0 , y 00 , · · · , y (n) when the independent variable is understood.
◦ Differential equations is a wide area of study and in order to facilitate the study and categorize
the solution methods several classifications can be introduced.
(i) Ordinary and partial differential equations
This is a main classification. An ordinary differential equation (ODE) is a differential equation in which the unknown function is a function of a single independent variable. A partial
differential equation (PDE) is a differential equation in which the unknown function is a function of more than one independent variable and the equation involves its partial derivatives. (In
this class we will study ODEs.)
Example: (i)(a) x2
(Here ν ∈ C.)
dy
d2 y
+ x + (x2 − ν 2 )y = 0. This is known as the Bessel’s equation. (An ODE.)
2
dx
dx
∂ 2 u(x, t)
∂u(x, t)
=
. This is known as the heat equation. (A
2
∂x
∂t
PDE.) (Here x is spatial variable, t is time variable and α is a positive constant.)
Example: (i)(b) α2 uxx = ut or α2
∂ 2 u(x, t)
∂ 2 u(x, t)
=
. This is known as the wave equation. (A
∂x2
∂t2
PDE.) (Here x is spatial variable, t is time variable and α is a positive constant.)
Example: (i)(c) α2 uxx = utt or α2
1
(ii) Order
The order of a differential equation is the order of the highest derivative that appears in the
equation.
dy
+ 5y − 1 = 0. A first order ODE.
dt
d2 y
dy
Example: (ii)(b) 2 − 3 + 4y = t3 . A second order ODE.
dt
dt
Example: (ii)(a) (1 + t2 )
Remark: F (t, y, y 0 , y 00 , · · · , y (n) ) = 0 is an nth order ODE and this is the general form of an nth
order ODE where F is a given function. However, most of the time we assume this can be solved
for y (n) and be written in the form y (n) = f (t, y, y 0 , y 00 , · · · , y (n−1) ).
(iii) Linear and nonlinear equations
This is a crucial classification. The ODE written in the form F (t, y, y 0 , y 00 , · · · , y (n) ) = 0 is called
linear if F is linear in y, y 0 , y 00 , · · · , y (n) . In other words, a linear ODE can be written in the form
0
an (t)y (n) (t) + an−1 (t)y (n−1) (t) + · · · + a1 (t)y (t) + a0 (t)y(t) = g(t).
If a differential equation is not linear it is called a nonlinear differential equation.
Example: (iii)(a) t2 y 00 + 3ty 0 − 2y = t3 + 1. (A linear ODE.)
g
d2 θ
Example: (iii)(b) 2 + sin θ = 0. (A nonlinear ODE-oscillating pendulum.) (Here g and L are
dt
L
constants.)
Remark: When modeling certain phenomena in terms of differential equations we may encounter
more than one unknown function. In such situations we arrive at a system of differential equations.
Example: (iv)

V
dI


=
dt
L
dV
1
V 

=− −
dt
C RC
(Here V is the voltage drop across the capacitor and I the current through the inductor in a LRC
parallel circuit and L is the inductance, C the capacitance, and R the resistance, constants.)
Remark: Once we are given a differential equation the most important questions are (1) is there
a solution to the given differential equation (existence), (2) if there is a solution is that the only
solution (uniqueness) and (3) how to find the solution (methodology).
Exercises 1. In each of Problems determine the order of the given differential equation and state
whether the equation is linear or nonlinear.
2
(a) (1 + t2 )y 000 + (sin t)y 0 − y = e−t
2d
2
dy 3
y
(b) x
− y = cos x
+ (cos x)
dx2
dx
(c) y (4) + y 00 + y 4 = 0.
(d) y (3) + (sin y)y 00 + y = sin t.
Exercises 2. In each of Problems verify that the given function(s) is a solution of the differential
equation.
(a) x
dy
− y = x2 ; y = x2 + 3x
dx
(b) y 00 + y = sec t (0 < t < π2 ) ; y = (cos t) ln cos t + t sin t
(c) y (4) + 4y (3) + 3y = x ; y1 = 31 x, y2 = e−x + 31 x
Z t
dy
2
2
t2
− 2ty = 1 ; y = e
e−r dr + et
(d)
dt
0
Exercises 3. Determine the values of r for which the differential equation y 000 − 3y 00 + 2y 0 = 0 has
solutions of the form y = ert .
Exercises 4. Determine the values of r for which the differential equation t2 y 00 + 4ty 0 + 2y = 0 has
solutions of the form y = tr where t > 0.
3
1.2 Direction fields
Exercises 5. (a) Suppose that an object is falling in the atmosphere near sea level. Formulate a
differential equation that describes the motion.
Suppose t denotes time (measured is seconds) and v (measured in ms−1 ) represents the velocity of
the falling object. Assume that v is positive in the downward direction. Apply Newton’s second
law (F = ma) where m is the mass of the object (measured in kg), a is acceleration (measured in
ms−2 and F is the net force exerted on the object (measured in N ). Note that gravity exerts a
force equal to the weight of the object, or mg, downward direction where g is the acceleration due
to gravity (assumed to be 9.8ms−2 ). A drag force due to air resistance is present and we assume it
has magnitude γv where γ is a constant called the drag coefficient (with units kgs−1 ). Thus, the
mathematical model becomes;
m
dv
= mg − γv.
dt
(b) Now suppose m = 10 and γ = 2. Then
v
dv
= 9.8 − .
dt
5
Investigate the behavior of solutions of this equation without solving the differential equation.
For selected values of v calculate dv
. For instance, if v = 40, dv
= 1.8. If v = 50, dv
= −.2. Solving
dt
dt
dt
dv
dv
=
0,
v
=
49.
Calculate
for
v
between
40
and
60
and
plot
the
small
line
segments
to a scale
dt
dt
for t between 0 to 10. This diagram is called a direction field or sometimes a slope field.
(c) Identify some solutions and comment on the behavior of solutions.
4
Exercises 6. Draw a direction field for differential equation y 0 = y + 2. Based on the direction
field, determine the behavior of y as t → ∞. If this behavior depends on the initial value of y at
t = 0, describe the dependency.
Exercises 7. Write down a differential equation of the form
y = 23 as t → ∞.
dy
dt
= ay + b whose solutions approach
Exercises 8. Draw a direction field for differential equation y 0 = y(y − 2)2 . Based on the direction
field, determine the behavior of y as t → ∞. If this behavior depends on the initial value of y at
t = 0, describe the dependency.
5
1.3 First order differential equations
1.3.1 First order linear differential equations
Suppose P, Q and R are arbitrary functions of t. A first order linear differential equation
can be written in the form
dy
(1)
P (t) + Q(t)y = R(t) (general form.)
dt
For P (t) 6= 0, taking q(t) =
Q(t)
P (t)
and r(t) =
R(t)
,
P (t)
equation (1) can be written in the form
dy
+ q(t)y = r(t)
dt
(standard form.)
(2)
Remark: An expression that contains all possible solutions of a given differential equation is known
as the general solution. The geometric representation of the general solution is an infinite family
of curves, known as integral curves. An additional data (such as the value of the solution at a
given value of the independent variable) is known as an initial condition. A differential equation
together with the initial condition form an initial value problem (IVP).
Exercises 9. Solve
(3 + t2 )
dy
+ 2ty = 4t.
dt
(3)
Exercises 10. Given that y(0) = 1 find the solution of
t
dy 1
+ 2 y = 12 e 3 .
(4)
dt
Exercises 11. By applying the method of integrating factor show that the general solution of
dy
+ ay = g(t) is given by
dt
Z t
y = e−at
eas g(s)ds + ce−at
t0
for some t0 < t where a is a given constant, g is a given function and c is an arbitrary constant.
Exercises 12. Find the general solution of
dy
− 2y = 4 − t.
dt
(5)
Using the method of integration factor, the general solution can be written as y = − 47 + 12 t + ce2t
where c is an arbitrary constant. Direction field
and some integral curves of (5) are shown below.
Observe the behavior of solution of (5) as t → ∞
for different values of c.
6
Exercises 13. By applying the method of integrating factor show that the general solution of
dy
+ p(t)y = g(t) is given by
dt
Z
i
1 h t
y=
µ(s)g(s)ds + c
µ(t) t0
for some t0 < t where c is an arbitrary constant and p and g are a given functions and µ(t) = e
(µ is known as the integrating factor ).
R
p(t) dt
.
Exercises 14. Solve the initial value problem ty 0 + 2y = 4t2 with y(1) = 2.
Using the method of integration factor, the required solution can be written as y = t2 + t12 . Direction field and some integral curves of ty 0 +2y =
4t2 are shown below. Investigate the behavior of
solution as t → 0 and t → ∞ for different values
of y(1).
Exercises 15. Solve the initial value problem 2y 0 + ty = 2 with y(0) = 1.
Using the method of integration factor, the required solution can be written as
Z t 2
2
s
t2
− t4
y=e
e 4 ds + e− 4 .
0
Direction field and some integral curves of ty 0 +
ty = 2 are shown below. For which value(s) y
approaches as t → ∞ for different values of y(0).
Exercises 16. Find the general solution of y 0 + y = 5 sin 2t.
Exercises 17. Find the solution of the initial value problem ty 0 + 2y = t2 − t + 1, y(1) = 21 , t > 0.
Exercises 18. Find the value of y0 for which the solution of the initial value problem y 0 − y =
1 + 3 sin t, y(0) = y0 remains finite as t → ∞.
7
1.3.2 Separable equations
• A differential equation in the form M (x) + N (y)
form) is known as a separable equation.
Exercises 19. Solve
dy
= 0 or M (x)dx + N (y)dy = 0 (differential
dx
dy
x2
=
.
dx
1 − y2
(6)
Exercises 20. Solve the initial value problem
3x2 + 4x + 2
dy
=
, y(0) = −1.
dx
2(y − 1)
⇒
y =1−
√
x3 + 2x2 + 2x + 4.
(7)
(8)
In this case, we are able to solve for y explicitly.
Note that solution exists only for x > −2 or interval of validity is (−2, ∞). Direction field
and integral curves of (7) are shown below. Note
the behavior of solution for the given nonlinear
equation.
Exercises 21. Solve
4x − x3
dy
=
dx
4 + y3
(9)
and determine the interval of validity of the solution which passes through the point (0,1).
⇒
x4 + y 4 − 8x2 + 16y − 17 = 0.
(10)
The interval of validity is (−3.35, 3.35) approximately. Direction field and integral curves of (9)
are shown below.
8
Exercises 22. Solve y 0 = cos2 x cos2 2y.
Exercises 23. Solve the initial value problem x dx + ye−x dy = 0, y(0) = 1.
Exercises 24. Solve
dy
x2 + 4xy + y 2
=
. (Use substitution y = xv(x).)
dx
x2
9
1.3.3 Modeling with first order differential equations
Exercises 25. At time t = 0 a tank contains Q0 lb of salt dissolved in 100 gal of water. Assume
that water containing 14 lb of salt/gal is entering the tank at a rate of r gal/min and that the
well-stirred mixture is draining from the tank at the same rate.
(a) Set up the initial value problem that describes this flow process.
(b) Find the amount of salt Q(t) in the tank at any time, and also find the limiting amount QL
that is present after a very long time.
(c) If r = 3 and Q0 = 2QL , find the time T after which the salt level is within 2% of QL .
(d) Also find the flow rate that is required if the value of T is not to exceed 45 min.
Exercises 26. Consider a pond that initially contains 10 million gal of fresh water. Water containing an undesirable chemical flows into the pond at the rate of 5 million gal/yr, and the mixture
in the pond flows out at the same rate. The concentration γ(t) of chemical in the incoming water
varies periodically with time according to the expression γ(t) = 2 + sin 2t g/gal. Construct a mathematical model of this flow process and determine the amount of chemical in the pond at any time.
Plot the solution and describe in words the effect of the variation in the incoming concentration.
Exercises 27. A body of constant mass m is projected away from the earth in a direction perpendicular to the earthy surface with an initial velocity v0 . Assuming that there is no air resistance, but
taking into account the variation of the earthy gravitational field with distance, find an expression
for the velocity during the ensuing motion. Also find the initial velocity that is required to lift
the body to a given maximum altitude ξ above the surface of the earth, and find the least initial
velocity for which the body will not return to the earth; the latter is the escape velocity.
Exercises 28. Suppose that a sum of money is deposited in a bank or money fund that pays interest at an annual rate r. The value S(t) of the investment at any time t depends on the frequency
with which interest is compounded for a given interest rate. Financial institutions have various
policies concerning compounding: some compound monthly, some weekly, some even daily.
(a) Find the amount in the account after 10 years if Rs. 1000 deposited at time t = 0 under the
annual rate of 8% and compounded m times per year for (i) m = 4 (ii) m = 12 (iii) m = 52 (iv)
m = 365. Describe the limiting behavior as m → ∞.
(b) If we assume that compounding takes place continuously, set up an initial value problem that
describes the growth of the investment and solve this equation.
10
1.3.4 Exact equations
Definition: Let
M (x, y) + N (x, y)y 0 = 0
(11)
be given. Suppose that there is a function ψ(x, y) such that
∂ψ
(x, y) = N (x, y),
∂y
∂ψ
(x, y) = M (x, y),
∂x
(12)
and such that ψ(x, y) = c (c constant) defines y = φ(x) implicitly as a differentiable function of x.
Then we say equation (11) is an exact differential equation.
Remark: In this case note that
M (x, y) + N (x, y)y 0 =
∂ψ ∂ψ dy
d
+
=
ψ(x, φ(x))
∂x
∂y dx
dx
and the equation (11) becomes
d
ψ(x, φ(x)) = 0
dx
and hence the solutions of (11) are given implicitly by ψ(x, y) = c where c is an arbitrary constant.
Theorem: Let the functions M, N, My , and Nx , where subscripts denote partial derivatives, be
continuous in the rectangular region R : α < x < β, γ < y < δ. Then equation (11)
M (x, y) + N (x, y)y 0 = 0
is an exact differential equation in R if and only if
My (x, y) = Nx (x, y)
at each point of R.
Proof: (Omitted)
Exercises 29. Solve 2x + y 2 + 2xyy 0 = 0.
Exercises 30. Solve (y cos x + 2xey ) + (sin x + x2 ey − 1)y 0 = 0.
Exercises 31. Solve 2x + 3 + (2y − 2)y 0 = 0.
Remark: It is sometimes possible to convert a differential equation that is not exact into an exact
equation by multiplying the equation by a suitable integrating factor.
Exercises 32. Determine whether (3xy + y 2 ) + (x2 + xy)y 0 = 0 is exact. (Attempt to solve this
equation after multiplying the equation by µ(x) = x.)
11
1.3.5 Existence and uniqueness
◦ So far,we have discussed a number of initial value problems, each of which had a solution and
apparently only one solution. That raises the question of whether this is true of all initial value
problems for first order equations. In other words, does every initial value problem have exactly
one solution? This may be an important question even for nonmathematicians. If you encounter
an initial value problem in the course of investigating some physical problem, you might want to
know that it has a solution before spending very much time and effort in trying to find it. Further,
if you are successful in finding one solution, you might be interested in knowing whether you should
continue a search for other possible solutions or whether you can be sure that there are no other
solutions.
Theorem 1: If the functions p and g are continuous on an open interval I : α < t < β containing
the point t = t0 , then there exists a unique function y = φ(t) that satisfies the differential equation
y 0 + p(t)y = g(t) for each t in I, and that also satisfies the initial condition y(t0 ) = y0 , where y0 is
an arbitrary prescribed initial value.
Proof: omitted.
be continuous in some rectangle α < t < β, γ < y < δ
Theorem 2: Let the functions f and ∂f
∂y
containing the point (t0 , y0 ). Then, in some interval t0 − h < t < t0 + h contained in α < t < β
there is a unique solution y = φ(t) of the initial value problem y 0 = f (t, y), y(t0 ) = y0 .
Proof: omitted.
Exercises 33. Find an interval in which the initial value problem ty 0 + 2y = 4t2 , y(1) = 2 has a
unique solution. (Apply Theorem 1.)
dy
3x2 + 4x + 2
=
, y(0) = −1. Apply Theorem 2 to the initial value
dx
2(y − 1)
problem and comment on existence and uniqueness of solution.
Exercises 34. Consider
dy
3x2 + 4x + 2
=
, y(0) = 1. Comment on the existence of solution of
dx
2(y − 1)
the initial value problem and applicability of Theorem 2.
Exercises 35. Consider
Exercises 36. Consider the initial value problem y 0 = y 1/3 , y(0) = 0 for t > 0. Apply Theorem 2
to this initial value problem and comment on existence and uniqueness of solution.
Exercises 37. Solve the initial value problem y 0 = y 2 , y(0) = 1 and determine the interval in
which the solution exists.
Exercises 38. Explain the differences between linear equations and nonlinear equations related to
existence of solution, uniqueness of solution, general solution, implicit and explicit solution.
Remark: If the equation is linear, the general solution can be written as an expression that contains
an arbitrary constant. In other words, the solution that contains an arbitrary constant contains all
possible solutions. However, this is not true for nonlinear equations.
12
1.4 Second order linear equations
◦ Let f be given. A second order ordinary differential equation has the form
y 00 (t) = f (t, y(t), y 0 (t)).
(13)
◦ If f is linear in y and y 0 , equation (13) is known as a linear equation. A second order linear
differential equation has the form
P (t)y 00 + Q(t)y 0 + R(t)y = G(t).
(14)
◦ If P (t) 6= 0, equation (14) has the form
y 00 + p(t)y 0 + q(t)y = g(t).
(15)
◦ If equation (13) is not linear, it is called a nonlinear equation.
◦ If g(t) = 0 for all t (in the domain of the equation), equation (15) is known as homogeneous
equation. If it is not homogeneous, it is known as nonhomogeneous equation. The term g in
equation (15) is known as nonhomogeneous term. Same terminology holds for equation (14).
◦ If P, Q and R are constant functions in equation (14), then it has the form
ay 00 + by 0 + cy = g(t).
(16)
◦ We will study solution methods to solve equation (16) and when solving equation (16) the first
step will be to solve the associated homogeneous equation
ay 00 + by 0 + cy = 0.
(17)
y 00 − y = 0.
(18)
Exercises 39. Solve
We seek a solution of the form of exponential function y = et . (Why?)
Show that y1 = et is a solution of equation (18).
Show that any constant multiple of y1 , c1 et is also a solution equation (18) where c1 is an arbitrary
constant.
Show that y2 = e−t is a solution of equation (18).
Show that any constant multiple of y2 , c2 e−t is also a solution equation (18) where c2 is an arbitrary
constant.
Show that sum of y1 and y2 is also a solution of equation (18).
Show that y = c1 et + c2 e−t (linear combinations of et and e−t ) is also a solution of equation (18).
Is there any other solution of equation (18) which we can’t obtain from y = c1 et + c2 e−t ?
◦ The solution y = c1 et + c2 e−t is called the general solution of equation (18) if we can obtain all
the solutions of equation (18) by assigning suitable values to arbitrary constants. (It is possible to
prove this. See notes later.)
Exercises 40. Solve the initial value problem (IVP)
y 00 − y = 0, y(0) = 2, y 0 (0) = −1.
(19)
Using the result in exercise 34, the general solution is y(t) = c1 et + c2 e−t . Show that the solution
of equation (19) is y = 21 et + 32 e−t by determining c1 and c2 .
13
Second order linear homogeneous differential equation with constant coefficients
(Solution method)
◦ Let a, b and c be given constants. Consider second order homogeneous linear differential equation
with constant coefficients;
ay 00 + by 0 + cy = 0.
(20)
◦ We seek exponential solutions of the form y = ert where r is a parameter to be determined.
Calculate y 0 and y 00 and substitute in equation (20) and obtain ert (ar2 + br + c = 0) = 0. Since
ert 6= 0,
ar2 + br + c = 0.
(21)
◦ Equation (21) is known as the characteristic equation or auxiliary equation for the differential equation (20).
◦ Note that if r is a root of the polynomial equation (21), then y = ert is a solution of the differential
equation (20).
◦ Three cases. (i) Equation (21) has two real distinct solutions (r1 6= r2 ). (ii) Equation (21)
has real repeated solutions (r1 = r2 ). (iii) Equation (21) has two complex (conjugate) solutions
(r1 , r2 = λ ± µi).
Case (i) : Characteristic equation ar2 + br + c = 0 has two real distinct solutions (r1 6= r2 )
In this case general solution of equation (24) is written as y(t) = c1 er1 t + c2 er2 t where c1 and c2 are
arbitrary constants.
Exercises 41. Find the general solution of y 00 + 5y 0 + 6y = 0. (answer. y = c1 e−2t + c2 e−3t .)
Exercises 42. Solve the initial value problem y 00 + 5y 0 + 6y = 0, y(0) = 2, y 0 (0) = 3. Comment on
the behavior of solution as t → ∞ (answer. y = 9e−2t − 7e−3t .)
Exercises 43. Solve the initial value problem 4y 00 − 8y 0 + 3y = 0, y(0) = 2, y 0 (0) = 21 . Comment
on the behavior of solution as t → ∞. (answer. y = − 12 e3t/2 + 52 et/2 .)
14
Second order linear differential equation with constant coefficients
◦ Let a, b and c be given constants. Consider second order homogeneous linear differential equation
with constant coefficients;
ay 00 + by 0 + cy = 0.
(22)
Case (ii) : Characteristic equation ar2 + br + c = 0 has repeated solutions (r1 = r2 = r)
In this case general solution of equation (22) is written as y(t) = c1 tert + c2 ert where c1 and c2 are
arbitrary constants.
Exercises 44. Solve the differential equation
y 00 + 4y 0 + 4y = 0.
(23)
Note that the characteristic equation is r2 + 4r + 4 = 0 ⇒ r = −2 and hence y1 = e−2t is one
solution. To find a second solution which is not a constant multiple of y1 we seek a solution of
the form y = v(t)y1 (t) where v(t) is a function of t to be determined. (This method is known as
reduction or order.) Calculate y 0 and y 00 and substitute in equation (23) and obtain v 00 (t) = 0.
Integrating twice obtain v(t) = c1 t + c2 . Thus y(t) = c1 te−2t + c2 e−2t .
Exercises 45. Find the solution of the initial value problem y 00 −y 0 +0.25y = 0, y(0) = 2, y 0 (0) = 31 .
Comment on the behavior of solution as t → ∞. (answer. y(t) = 2et/2 − 23 tet/2 .)
Exercises 46. Find the solution of the initial value problem y 00 −y 0 +0.25y = 0, y(0) = 2, y 0 (0) = 2.
Comment on the behavior of solution as t → ∞. (answer. y(t) = 2et/2 + tet/2 .)
Exercises 47. (Method of reduction of order ) Given that y1 (t) = t−1 is a solution of the
differential equation 2t2 y 00 + 3ty 0 − y = 0, t > 0 find the general solution.
Set y(t) = t−1 v(t). Calculate y 0 and y 00 and substitute in the given equation and obtain 2tv 00 −v 0 = 0.
Now let w = v 0 and arrive at 2tw0 − w = 0. Solve this for w(t) = ct1/2 and find v(t) = 23 ct1/2 + k
and thus y(t) = 32 ct1/2 + kt−1 where c and k are arbitrary constants.
15
Second order linear differential equation with constant coefficients
◦ Let a, b and c be given constants. Consider second order homogeneous linear differential equation
with constant coefficients;
ay 00 + by 0 + cy = 0.
(24)
Case (iii) : Characteristic equation ar2 + br + c = 0 has two complex solutions
(r1 = λ + µi, r2 = λ − µi) (λ, µ ∈ R, µ 6= 0)
In this case general solution of equation (24) is written as y(t) = c1 eλt cos µt + c2 eλt sin µt where c1
and c2 are arbitrary constants.
Remark: Euler’s formula: eiµt = cos µt + i sin µt where µ and t are real numbers.
Remark: If r1 = λ + µi, then y1 = er1 t = e(λ+µi)t = eλt eµit = eλt (cos µt + i sin µt) = eλt cos µt +
ieλt sin µt. (Similarly, if r2 = λ − µi, then y2 = eλt cos µt − ieλt sin µt.) In either case real part is
eλt cos µt and the imaginary part is eλt sin µt and in this case real and imaginary parts form a fundamental set of solutions. Thus the general solution can be written as y(t) = c1 eλt cos µt + c2 eλt sin µt.
(It is possible to prove this. See notes later.)
Exercises 48. Solve the initial value problem y 00 + y 0 + 9.25y = 0, y(0) = 2, y 0 (0) = 8. Comment
on the behavior of solution as t → ∞. (answer. y = e−t/2 (2 cos 3t + 3 sin 3t)).
Exercises 49. Find the solution of the initial value problem 16y 00 − 8y 0 + 145y = 0, y(0) =
−2, y 0 (0) = 1. Comment on the behavior of solution as t → ∞. (answer. y = et/4 (−2 cos 3t +
1
sin 3t)).
2
Exercises 50. Find the general solution of y 00 + 9y = 0. Comment on the behavior of solution as
t → ∞. (answer. y = c1 cos 3t + c2 sin 3t).
16
*1.5 Second order linear equations (Theorems) (*FYI)
Theorem 1: (Existence and uniqueness Theorem)
Consider the initial value problem
y 00 + p(t)y 0 + q(t)y = g(t) y(t0 ) = y0 , y 0 (t0 ) = y00
(25)
where p, q, and g are continuous on an open interval I that contains the point t0 . Then there is
exactly one solution y = φ(t) of this problem, and the solution exists throughout the interval I.
Proof: Omitted.
Exercises 51. Consider the initial value problem (t2 −3t)y 00 +ty 0 −(t+3)y = 0, y(1) = 2, y 0 (1) = 1.
Find the longest interval in which the solution is certain to exist.
Exercises 52. Find the unique solution of the initial value problem y 00 + p(t)y 0 + q(t)y = 0, y(t0 ) =
0, y 0 (t0 ) = 0 where p and q are continuous in an open interval I containing t0 .
Theorem 2: (Principle of superposition)
If y1 and y2 are two solutions of the differential equation y 00 + p(t)y 0 + q(t)y = 0 then the linear
combination c1 y1 + c2 y2 is also a solution for any values of the constants c1 and c2 .
Proof: Omitted.
Theorem 3: (Wronskian)
Suppose that y1 and y2 are two solutions of y 00 + p(t)y 0 + q(t)y = 0, and that the initial conditions
y(t0 ) = y0 and y 0 (t0 ) = y00 are assigned. Then it is always possible to choose the constants c1 , c2 so
that y = c1 y1 + c2 y2 satisfies the differential equation and the initial conditions if and only if the
Wronskian W = y1 y20 − y10 y2 6= 0 at t = t0 .
Proof: Omitted.
Exercises 53. Solve y 00 + 5y 0 + 6y = 0 and find the Wronskian W (y1 , y2 )(t).
Theorem 4: (General solution)
Suppose that y1 and y2 are two solutions of y 00 + p(t)y 0 + q(t)y = 0. Then the family of solutions
y = c1 y1 + c2 y2 with arbitrary coefficients c1 , c2 includes every solution if and only if there is a point
t0 where the Wronskian of y1 and y2 is not zero.
Proof: Omitted.
17
Remark: The expression y = c1 y1 + c2 y2 with arbitrary constant coefficients is known as the general solution of y 00 + p(t)y 0 + q(t)y = 0. The solutions y1 and y2 are said to form a fundamental
set of solutions of y 00 + p(t)y 0 + q(t)y = 0 if and only if their Wronskian is nonzero.
Exercises 54. Suppose that y1 (t) = er1 t and y2 (t) = er2 t are two solutions of the differential
equation y 00 + p(t)y 0 + q(t)y = 0. Show that they form a fundamental set of solutions if r1 6= r2 .
Exercises 55. Show that y1 (t) = t1/2 and y2 (t) = t−1 form a fundamental set of solutions of
2t2 y 00 + 3ty 0 − y = 0, t > 0.
Theorem 5: (Fundamental set of solutions)
Consider the differential equation y 00 + p(t)y 0 + q(t)y = 0, (1) whose coefficients p and q are continuous on some open interval I. Choose some point t0 in I. Let y1 be the solution of equation (1)
that also satisfies the initial conditions y(t0 ) = 1, y 0 (t0 ) = 0, and let y2 be the solution of equation
(1) that satisfies the initial conditions y(t0 ) = 0, y 0 (t0 ) = 1. Then y1 and y2 form a fundamental set
of solutions of equation (1).
Proof: Omitted.
Exercises 56. Find the fundamental set of solutions y1 and y2 specified by Theorem 5 for the
differential equation y 00 − y = 0, using initial point t0 = 0.
Theorem 6: (Complex solutions)
Consider the differential equation y 00 + p(t)y 0 + q(t)y = 0, (1) where p and q are continuous realvalued functions. If y = u(t) + iv(t) is a complex valued solution of equation (1) then its real part
u and its imaginary part v are also solutions of this equation.
Proof: Omitted.
Theorem 7: (Abel’s formula)
If y1 and y2 are solutions of the differential equation y 00 + p(t)y 0 + q(t)y = 0, (1) where p and q are
continuous on an open interval I, then the Wronskian W (y1 , y2 )(t) is given by
h Z
i
W (y1 , y2 )(t) = c exp − p(t)dt
where c is a certain constant that depends on y1 and y2 , but not on t. Further, W (y1 , y2 )(t) either
is zero for all t in I (if c = 0) or else is never zero in I (if c 6= 0).
Proof: Omitted.
Exercises 57.
Consider 2t2 y 00 + 3ty 0 − y = 0, t > 0. Calculate Wronskian W .
18
1.6 Nonhomogeneous equations; method of undetermined coefficients
Note: Let p and q be continuous functions on an open interval I and φ be a differentiable function.
We may define a differential operator L by the equation
L[φ] = φ00 + pφ0 + qφ.
The value of L[φ] at a point t ∈ I is
L[φ](t) = φ00 (t) + p(t)φ0 (t) + q(t)φ(t).
The operator L can also be written as L = D2 + pD + q, where D =
d
dt
is the derivative operator.
◦ Let p and q be continuous functions on an open interval I. Consider the nonhomogeneous
equation
L[y] := y 00 + p(t)y 0 + q(t)y = g(t).
(26)
The equation
L[y] := y 00 + p(t)y 0 + q(t)y = 0.
(27)
in which g(t) = 0 and p, q are the same as in (26), is called the homogeneous equation corresponding
to (26).
Theorem: If Y1 and Y2 are two solutions of the nonhomogeneous equation (26), then their difference Y1 − Y2 is a solution of the corresponding homogeneous equation (27). If, in addition, y1 and
y2 are a fundamental set of solutions of (27), then Y1 (t) − Y2 (t) = c1 y1 (t) + c2 y2 (t), where c1 and c2
are certain constants.
Proof: Omitted.
Theorem: The general solution of the nonhomogeneous equation (26) can be written in the form
y = φ(t) = c1 y1 (t) + c2 y2 (t) + Y (t), where y1 and y2 are a fundamental set of solutions of the
corresponding homogeneous equation (27), c1 and c2 are arbitrary constants, and Y is some specific
solution of the nonhomogeneous equation (26).
Proof: Omitted.
◦ To solve the nonhomogeneous equation (26), we must do three things:
1. Find the general solution c1 y1 (t) + c2 y2 (t) of the corresponding homogeneous equation. This
solution is frequently called the complementary solution and may be denoted by yc (t).
2. Find some single solution Y (t) of the nonhomogeneous equation. Often this solution is referred
to as a particular solution.
3. Form the sum of the functions found in steps 1 and 2.
◦ To find a particular solution Y (t) of the nonhomogeneous equation (26) we use the method of
undetermined coefficients.
19
Exercises 58. Find a particular solution of y 00 − 3y 0 − 4y = 3e2t . (Assume Y (t) = Ae2t and find
A. Ans. Y (t) = − 12 e−2t .)
Exercises 59. Find a particular solution of y 00 − 3y 0 − 4y = 2 sin t. (Assume Y (t) = A sin t + B cos t
5
3
and find A, B. Ans. Y (t) = − 17
sin t + 17
cos t.)
Exercises 60. Find a particular solution of y 00 − 4y 0 − 12y = 2t3 − t + 3. (Assume Y (t) =
5
At3 + Bt2 + Ct + D and find A, B, C, D. Ans. Y (t) = − 16 t3 + 16 t2 − 19 t − 27
.)
Exercises 61. Find a particular solution of y 00 −3y 0 −4y = −8et cos 2t. (Assume Y (t) = Aet cos 2t+
2 t
10 t
e cos 2t + 13
e sin 2t.)
Bet sin 2t and find A, B. Ans. Y (t) = 13
Remark: Suppose that g(t) is the sum of two terms, g(t) = g1 (t) + g2 (t), and suppose that Y1 and
Y2 are solutions of the equations ay 00 + by 0 + cy = g1 (t) and ay 00 + by 0 + cy = g2 (t), respectively.
Then Y1 + Y2 is a solution of the equation ay 00 + by 0 + cy = g(t).
Exercises 62. Find a particular solution of y 00 − 3y 0 − 4y = 3e2t + 2 sin t − 8et cos 2t.
3
5
10 t
2 t
cos t − 17
sin t + 13
e cos 2t + 13
e sin 2t.)
(Ans. Y (t) = − 21 e2t + 17
Exercises 63. Find a particular solution of y 00 − 3y 0 − 4y = 2e−t . (Assume Y (t) = Ate−t and find
A. Ans. Y (t) = − 52 te−t .)
Exercises 64. Find the general solution of y 00 + ω02 y = cos ω0 t. (ω0 constant)
Exercises 65. Solve the initial value problem y 00 − 2y 0 − 3y = 3te2t , y(0) = 1, y 0 (0) = 0.
20
1.7 Applications of second order equations
◦ Second order linear equations with constant coefcients finds applications in diverse areas. For
instance, the motion of a mass on a vibrating spring, and the ow of electric current in a simple
series circuit are modeled and described using second order equations.
1.7.1 Spring mass systems
Let m be the mass of the object, γ the damping
constant, k the spring constant, F a given external
force, and u the displacement measured from the
equilibrium position. (m, γ, k positive constants.)
Exercises 66. Applying Newton’s law, Hooke’s law and any other assumptions show that the
motion of the object can be modeled by the equation mu00 + γu0 + ku = F .
Exercises 67. A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is displaced an
additional 6 in. in the positive direction and then released. The mass is in a medium that exerts
a viscous resistance of 6 lb when the mass has a velocity of 3 ft/sec. Formulate the initial value
problem that governs the motion of the mass. (State any assumptions you made.)
Exercises 68. If the damping constant γ = 0 and the external force F (t) ≡ 0 then the motion is
known as undamped free vibration and the related equation is mu00 + ku = 0.
(a) Show that the solution is given by u = A cos ω0 t + B sin ω0 t where A and B are arbitrary conk
.
stants and ω02 = m
(b) Show that the solution can also be written as u = R cos(ω0 t − δ) where R =
.
tan δ = B
A
√
A2 + B 2 and
(c) Identify the frequency, period, amplitude, and the phase of this motion.
Exercises 69. A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its
equilibrium position with a downward velocity of 10 cm/sec, and if there is no damping, determine
the position u of the mass at any time t. When does the mass rst return to its equilibrium position?
Exercises 70. Suppose the motion of a spring mass system is governed by the equation mu00 +ku =
F0 cos ωt where m, k, F0 are constants. Show that the solution is given by
F0
u = A cos ω0 t + B sin ω0 t + 2
cos ωt
ω0 − ω 2
q
k
where ω0 = m
6= ω.
21
1.7.2 Electric circuits
Here the current I (amperes), is a function of time
t. The resistance R (ohms), the capacitance C
(farads), and the inductance L (henrys) are all
positive and are assumed to be known constants.
The impressed voltage E (volts) is a given function of time. The total charge Q (coulombs) on
the capacitor at time t.
Exercises 71. (a) Using elementary laws in electricity and Kirchhoff ’s law show that
1
dI
L + RI + Q = E(t).
dt
C
1
(b) Show that LQ00 + RQ0 + Q = E(t).
C
1
(c) Show that LI 00 + RI + I = E 0 (t).
C
(d) Comment on the initial conditions.
22
1.8 Numerical methods
◦ As there are many differential equations arising in applications that cannot be solved analytically
we use numerical methods to approximate solutions of such equations.
◦ Let t ∈ [a, b]. Consider the initial value problem (IVP)

dy
= f (t, y)
dt

y(0) = y0 .
We refer yn as an approximation to the exact value y(tn ) where tn = nh, h =
(28)
b−a
,
n
and n ≥ 1.
1.7.1 Eulers method (explicit method)
◦ We approximate the derivative in (28) using a forward difference formula
f (t, y) ≈
y(t + h) − y(t)
.
h
We denote f (nh, yn ) by fn . Then, the Eulers method is given by;
yn = yn−1 + hfn−1 .
Remark: This is an explicit method.
Exercises 72. Let

1 
dy
=−
dt
(t + y)2

y(0) = 0.9.
Carry out two time steps of Euler’s method with a step size of h = 0.125 so as to obtain approximations to y(0.125) and y(0.25). Give your answer to 6 decimal places.
Exercises 73. Let

dy
2
= −y
dt

y(0) = 0.5.
Carry out two time steps of Euler’s method with a step size of h = 0.01 so as to obtain approximations to y(0.01) and y(0.02). Give your answer to 6 decimal places.
Exercises 74. Let

dy
=y
dt

y(0) = 1.
Approximate y(1) to 6 decimal places using Euler’s method with a step size of (a) h = 0.2 (b)
h = 0.1 (c) h = 0.05 (d) h = 0.025 (e) h = 0.0125. Compare your answer with exact answer (by
calculating difference.)
23
Remark: The error |y(tn ) − yn | in Euler’s method is (approximately) proportional to the step size
h. This sort of behavior is called first-order. Thus, Euler’s method is first order. In other words,
the error it incurs is approximately proportional to h.
1.7.2 Eulers method (implicit method)
◦ Let t ∈ [a, b]. Consider the initial value problem (IVP)

dy
= f (t, y)
dt

y(0) = y0 .
We refer yn as an approximation to the exact value y(tn ) where tn = nh, h =
(29)
b−a
,
n
and n ≥ 1.
Note that by integrating (30) from t = tn to t = tn+1 we get
Z tn+1
yn+1 − yn =
f (t, y) dt.
tn
Now approximating the integral by trapezoidal rule we get
h
yn+1 = yn + (fn + fn+1 ).
2
Remark: This is an implicit method. This is also known as trapezium method.
Exercises 75. Let

dy
1 
=
dt
1+t

y(0) = 1.
Carry out two time steps of trapezium method with a step size of h = 0.2 so as to obtain approximations to y(0.2) and y(0.4). Give your answer to 6 decimal places.
Exercises 76. Let

dy
1
=
− 2y 
dt
1 + t2

y(0) = 2.
Carry out two time steps of trapezium method with a step size of h = 0.1 so as to obtain approximations to y(0.1) and y(0.2). Give your answer to 6 decimal places.
Exercises 77. Let

dy
=y
dt

y(0) = 1.
Approximate y(1) to 6 decimal places using trapezium method with a step size of (a) h = 0.2 (b)
h = 0.1 (c) h = 0.05 (d) h = 0.025 (e) h = 0.0125. Compare your answer with exact answer (by
calculating difference.)
24
Remark: The error |y(tn ) − yn | in trapezium method is (approximately) proportional to the step
size h2 . This sort of behavior is called second-order. Thus, trapezium method is second order. In
other words, the error it incurs is approximately proportional to h2 .
1.7.3 The Runge-Kutta method (RK4)
◦ Let t ∈ [a, b]. Consider the initial value problem (IVP)

dy
= f (t, y)
dt

y(0) = y0 .
We refer yn as an approximation to the exact value y(tn ) where tn = nh, h =
(30)
b−a
,
n
and n ≥ 1.
The Runge-Kutta method is given as:
h
yn+1 = yn + (kn1 + 2kn2 + 2kn3 + kn4 )
6
where
kn1
kn2
kn3
kn4
= f (tn , yn ),
= f (tn + 21 h, yn + 12 hkn1 ),
= f (tn + 12 h, yn + 12 hkn2 ),
= f (tn + h, yn + hkn3 ).
Exercises 78. Let

dy
= 1 − t + 4y 
dt

y(0) = 1.
Approximate y(1) to 6 decimal places using Runge-Kutta method with a step size of (a) h = 0.2
(b) h = 0.1 (c) h = 0.05 (d) h = 0.025 (e) h = 0.0125. Compare your answer with exact answer
(by calculating difference.)
Remark: This method has a local truncation error that is proportional to h5 . This is a forth order
method.
25
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