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LINEAR ODES OF HIGHER ORDER

Stephenn L. Rabano, PECE

Second-Order Linear ODEs

Recall that ODEs may be classified as linear or nonlinear.

Nonlinear ODEs of second or higher order are generally difficult to solve.

Linear ODEs are less difficult that various properties of their solutions can be generalized and standard methods are available to solve some of them.


Linear ODEs of the second order are most important in mechanical and electrical engineering applications.

Applications include mechanical and electrical vibrations, wave motion, heat conduction, and other physical phenomena.

Second Order Homogeneous Linear ODEs A linear second-order ODE is written

y + p(x) y + q(x) y = r(x)

A nonlinear second-order ODE is written otherwise.

Note that the previous equation shown is linear in y and its derivatives, and p, q, and r may be any given function of x.

It is the standard form of linear second-order ODEs.


If r(x) is identically zero (r(x) = 0 for all x considered), the equation becomes the standard form of a homogeneous linear second order ODE.

Otherwise, it is nonhomogeneous.

Examples: Classify each ODE according to linearity and homogeneity (if linear). y + 25y = e-x cos x

xy + y + xy = 0

y y + (y)2 = 0


p and q in the standard form of linear second-order ODEs are known as coefficients of the ODEs.

y = h(x) is a solution of either a linear or a nonlinear second-order ODE on some open interval I if h is defined and twice differentiable throughout I and such that the ODE becomes an identity if y is replaced by h, y by h, and y by h.

Superposition Principle for Homogeneous Linear ODEs Superposition principle or linearity principle

states that further solutions from given ones can be obtained by adding them or by multiplying them with any constant, an advantage of homogeneous linear ODEs.

Example: Verify that y = cos x and y = sin x for all x are solutions of the homogeneous linear ODE

y + y = 0

Superposition Principle for Homogeneous Linear ODEs Note also that multiplying cos x by any

constant, say 4.7, and sin x by any other constant, say -2, and taking the sum of the results gives another solution of the previous example.

Generally, y = c1y1 + c2y2 where c1 and c2 are arbitrary constants.

The previous function (expression on the right side of the previous equation) is called a linear combination of y1 and y2.

Superposition Principle for Homogeneous Linear ODEs Theorem: For a homogeneous linear ODE,

any linear combination of two solutions on an open interval I is again a solution of the ODE on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.

Example: Verify that y = c1y1 + c2y2 is a solution of y + py + qy = 0 if each of y1 and y2 is a solution of the said ODE.

Superposition Principle for Homogeneous Linear ODEs Example: Verify that the functions y = 1 + cos

x and y = 1 + sin x are solutions of the nonhomogeneous linear ODE y + y = 1 and their sum, 2(1 + cos x), 5(1+ sin x) are not solutions.

Example: Verify that the functions y = x2 and y = 1 are solutions of the nonlinear ODE yy xy = 0 but their sum and x2 are not solutions.

Initial Value Problem: Second-Order Homogeneous ODE An initial value problem for a second-order

homogeneous linear ODE has two initial conditions:

y(x0) = K0 and y(x0) = K1 The said conditions prescribe given values of

K0 and K1 of the solution and its first derivative (slope of its curve) at the same given x = x0 in the considered open interval.

Initial Value Problem: Second-Order Homogeneous ODE Additionally, they are used to determine two

arbitrary constants c1 and c2 in a general solution y = c1y1 + c2y2 of the ODE.

y1 and y2 are suitable solutions of the ODE, resulting in a unique solution (particular solution of the ODE) passing through the point (x0, K0) with K1 as the tangent direction (slope) at that point.

Initial Value Problem: Second-Order Homogeneous ODE Example: Solve the initial value problem y +

y = 0, y(0) = 3, y(0) = -0.5.

Initial Value Problem: Second-Order Homogeneous ODE If two proportional solutions y1 = cos x and y2

= k cos x are introduced so that y1/y2 = 1/k = constant, then y = c1y1 + c2y2 can be written in the form y = c1 cos x + c2(k cos x) = c cos x where c = c1 + c2k.

Since the two initial conditions will no longer be satisfied with only one arbitrary constant c, proportionality must be excluded in defining the concept of a general solution.

Initial Value Problem: Second-Order Homogeneous ODE Generally, a general solution of a second-

order homogeneous ODE on an open interval I is a solution in which y1 and y2 are solutions of the ODE on I that are not proportional, and c1 and c2 are arbitrary constants.

y1 and y2 are called a basis (or a fundamental system) of solutions of the ODE on I.

A particular solution of the ODE on I is obtained by assigning specific values to c1 and c2.

Initial Value Problem: Second-Order Homogeneous ODE A basis may also be defined using the

concept of linear dependence.

Functions y1 and y2 are called linearly independent on an interval I where those functions are defined if k1y1(x) + k2y2(x) = 0 everywhere on I implies k1 = 0 and k2 = 0.

Otherwise, if k1y1(x) + k2y2(x) = 0 holds for some constants k1 and k2 not both zero, y1 and y2 are called linearly dependent.

Initial Value Problem: Second-Order Homogeneous ODE If k1 0 or k2 0, y1 and y2 are proportional.

If the two functions are linearly independent, they are not proportional.

A basis of solutions of the ODE on an open interval I, therefore, is a pair of linearly independent solutions of the ODE on I.

If the coefficients p and q are continuous on some open interval I, then the ODE has a general solution where it yields the particular solution of any initial value problem ODE.

Initial Value Problem: Second-Order Homogeneous ODE It also includes all solutions of the ODE on I

and no singular solutions (solutions not obtainable from a general solution) exist.

Example: From the previous problem cos x and sin x form a basis of solutions of the ODE y + y = 0 for all x because their quotient (cot x or tan x) is not constant (and they are said to be not proportional).

Initial Value Problem: Second-Order Homogeneous ODE Example: Verify that y1 = e

x and y2 = e-x are

solutions of y y = 0. Solve the initial value problem y y = 0, y(0) = 6, y(0) = -2. Hint: y1 and y2 must form a basis for all x to solve the initial value problem.

Reduction of Order: Homogeneous ODE of Order 2 There are cases when one solution (one of

two linearly independent solutions) can be found by inspection or in some other way.

A second linearly independent solution can be obtained by solving a first-order ODE called the method of reduction of order.

For example, in finding a basis of solutions of (x2 x)y xy + y = 0, note that by inspection, y1 = x is a solution.

Reduction of Order: Homogeneous ODE of Order 2 The method suggests that substitutions

based on the following must be done

y2 = uy1 = ux, y = ux + u, y = ux + 2u

giving (x2 x)(ux + 2u) x(ux + u) + ux = 0.

Simplifying and dividing by x:

(x2 x)u + (x 2)u = 0

Making the ODE a first order ODE in v = u, the ODE may now be solved by method of separating variables.

Homogeneous Linear ODEs with Constant Coefficients Consider a second-order homogeneous linear

ODEs whose coefficients a and b are constant:

y + ay + by = 0

Applications of these equations include mechanical and electrical vibrations.

Recall that y = ce-kx is the solution of the first-order linear ODE (with constant coefficient k) y + ky = 0.

Homogeneous Linear ODEs with Constant Coefficients Try the solution y = ex and find its derivatives.

Substitute to the ODE to obtain the characteristic equation (auxiliary equation) (obtained by dividing the resulting ODE by ex.

The ODE then becomes (2 + a + b) ex = 0.

The characteristic equation then is the equation (2 + a + b) = 0.

Homogeneous Linear ODEs with Constant Coefficients If is a solution of the important

characteristic equation, then the exponential function ex is a solution of the ODE.

Note that the quadratic characteristic equation has the following roots from the quadratic formula of elementary algebra:

1 = ()[-a + (a2 4b)1/2]

2 = ()[-a (a2 4b)1/2]

Homogeneous Linear ODEs with Constant Coefficients The derivations shows that the functions y1 =

e1x and y2 = e2x are solutions of the ODE

which when verified are really valid particular solutions.

From algebra as well, there are three kinds of roots depending on the sign of the discriminant a2 4b:

Case 1: Two real roots if a2 4b > 0.

Case 2: A real double root if a2 4b = 0.

Homogeneous Linear ODEs with Constant Coefficients

Case 3: Complex conjugate roots if a2 4b < 0.

For two distinct real roots (Case 1), a basis of solution of the ODE on any interval is y1 = e

1x and y2 = e

2x since y1 and y2 are defined (and real) for all x and their quotient is not constant.

The corresponding general solution is:

y = c1 e1x + c2 e

2x

Homogeneous Linear ODEs with Constant Coefficients Example: Solve the following ODE.

y y = 0

y + y 2y = 0, y(0) = 4, y(0) = 5

For Case 2 (real double root), there is only one root = 1 = 2 = a/2.

Therefore, there is only one solution

y1 = e(a/2)x

Homogeneous Linear ODEs with Constant Coefficients The method of reduction of order may be

used to obtain a second independent solution y2.

First, set y = uy1 and substitute this and its derivatives y and y into the standard form.

Collect terms in u, u, and u to rewrite the equation.

Since y1 is a solution of the ODE, the coefficient of u is zero.

Homogeneous Linear ODEs with Constant Coefficients The coefficient of u is also zero since:

2y = ae(a/2)x = ay1 The previous findings will give the resulting

equation uy1 = 0 and dividing both sides by y1 gives u = 0.

Integrating both sides of the last equation with respect to x, it yields u = c1x + c2.

For the second independent solution y2 = uy1, c1 = 1 and c2 = 0 can be chosen so that u = x.

Homogeneous Linear ODEs with Constant Coefficients Therefore, y2 = xy1, and obviously, these

solutions are not proportional and thus form a basis.

A basis of solutions for Case 2 on any interval is e-ax/2, xe-ax/2.

The general solution then will be

y = (c1 + c2x) e-ax/2

Homogeneous Linear ODEs with Constant Coefficients Example: Solve the following.

y + 6y + 9y = 0

y + y + 0.25y = 0, y(0) = 3, y(0) = 3.5

For Case 3, the magnitude of the imaginary number of each root is given by the following:

1

22 4 =

1

2(4 2) =

2

4= j

2

4

Homogeneous Linear ODEs with Constant Coefficients The roots therefore are:

1=

2+ j

2

4

2=

2 j

2

4

The solutions therefore are:

e1x =

2+

2

4=

2(cos

2

4 +

sin 2

4)

Homogeneous Linear ODEs with Constant Coefficients

e2x =

2

2

4=

2(cos

2

4

sin 2

4)

Adding the two solutions and multiplying the

result by give 1 =

2 cos

2

4.

Homogeneous Linear ODEs with Constant Coefficients Subtracting the second solution from the first

and dividing the result by j2 give

2 = 2 sin

2

4

The results are again solutions since they are obtained by addition and multiplication by constants based on the superposition principle.

Homogeneous Linear ODEs with Constant Coefficients Note that the two previous solutions are real

and therefore, a real general solution in Case 3 is =

2 1 cos

2

4 + 2sin

2

4

Example: Solve the following equations.

y + 0.4y + 9.04y = 0, y(0) = 0, y(0) = 3

y + 2y = 0

Modeling Electric Circuits

Linear ODEs are models of electric circuits, as they occur as part of large networks in computers and other systems.

The circuit shown in the next slide has 3 components: resistor (in ohms), inductor (in henrys), and capacitor (in farads).

Such is called an RLC-circuit.

All components are wired in series and connected to an electromotive force E(t) (which may be a generator) in volts.


R, L, C, and E are given to find the current I(t) A (amperes) in the circuit.


An ODE for the current of the previous circuit is obtained from Kirchhoffs Voltage Law (KVL): The voltage (electromotive force) impressed on a closed loop is equal to the sum of the voltage drops across the other elements of the loop.

From KVL, the voltage drop (caused by a current I flowing through the components) across each component is given by the following:


RI from Ohms law, voltage drop for a resistor of resistance R ohms ()

LI = L dI/dt voltage drop for an inductor of inductance L henrys (H)

Q/C voltage drop for a capacitor of capacitance C farads (F)

And since Q coulombs is the charge on the capacitor related to the current by I(t) = dQ/dt such that Q(t) = I(t) dt, the model for the previous circuit is the integro-differential equation

01

LI' RI Idt E t E sin tC


Differentiating the equation with respect to t to avoid the integral gives:

The current then in an RLC-circuit is obtained as the solution of the previous equation which is a nonhomogenous second-order ODE with constant coefficients.

01

LI'' RI' I E' t E cos tC


Similarly, since I = Q and I = Q, the equation below has Q as its unknown.

01

LQ'' RQ' Q E t E sin tC

COMPLEX NUMBERS (Z)


z = x + jy

Historically, equations without real solutions were observed and led to the introduction of complex numbers.

Examples: x2 = 1, x2 10x + 40 = 0

A complex number z is an ordered pair (x, y) of real numbers x and y:

z = (x, y)

z = x + jy

x is called the real part of z:

x = Re z

y is called the imaginary part of z:

y = Im z

Definition: Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

z = x + jy

(0, 1) is called the imaginary unit and is denoted by i:

i = (0, 1)

Basic operations on complex numbers:

Addition

Multiplication

Subtraction

Division

z = x + jy

C0mplex plane: Cartesian coordinate system

real axis, x-axis

imaginary axis, y-axis

z = x + jy

Example: 4 j3 in the complex plane

z = x + jy

Definition: The xy-plane in which complex numbers are represented is called the complex plane.

The complex conjugate of a complex number z = x + jy is defined by

It is obtained geometrically by reflecting the point z in the real axis.

z = x + jy

z = x + jy

Complex numbers follow the following:

commutative laws of addition and multiplication

associative laws of addition and multiplication

distributive law of multiplication over addition

z = x + jy

Addition of complex numbers:

z = x + jy

Subtraction of complex numbers:

z = x + jy

Examples:

Evaluate the reciprocal of the imaginary unit.

Evaluate the successive integral powers of the imaginary unit.

Polar Form

The complex plane is also useful in obtaining the polar form of complex numbers using the polar coordinates (r, ).

Notice that x = r cos and y = r sin .

z = x + jy takes the polar form

z = r(cos + j sin ).

r is called the absolute value or modulus of z denoted by |z|.

Polar Form

is called the argument of z and is denoted by arg z with the principal value Arg z:

< Arg z

Note that the double inequality specifies the value of the argument generated by standard scientific calculators.

is the directed angle from the positive x-axis.

Polar Form

Note that it is measured positive in the counterclockwise sense.

For z = 0, the argument is undefined.

For a given z 0, it is determined only up to integer multiples of 2 since cosine and sine are periodic with period 2.

The other values of arg z are arg z = Arg z 2n (n = 1, 2, ).

Polar Form

Polar Form

Distance between two points in the complex plane:

Polar Form

Express 1 + j in its polar form.

COMPLEX NUMBERS (Z)


z = r cos + j r sin

z = x + j y

Triangle Inequality

z = x + j y

z = r cos + j r sin

z = rej

LAPLACE AND INVERSE LAPLACE TRANSFORMS


Laplace Transform

MATRICES AND DETERMINANTS


Matrix

Determinant

mathece.ppt's.pdf

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