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Solve the following initial-value problem and graph the solution: y 2 y + 10 y = 0 , y ( 0 ) = 2 , y ( 0 ) = −1

y ( x ) = e x ( 2 cos 3 x sin 3 x )
This figure is the graph of y(x) = e^x(2 cos 3x − sin 3x) It has the positive x axis scaled in increments of even tenths. The y axis is scaled in increments of twenty. The graph itself starts at the origin. Its amplitude increases as x increases.

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Initial-value problem representing a spring-mass system

The following initial-value problem models the position of an object with mass attached to a spring. Spring-mass systems are examined in detail in Applications . The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. (Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.)

y + 2 y + y = 0 , y ( 0 ) = 1 , y ( 0 ) = 0

Solve the initial-value problem and graph the solution. What is the position of the mass at time t = 2 sec? How fast is the mass moving at time t = 1 sec? In what direction?

In [link] c. we found the general solution to this differential equation to be

y ( t ) = c 1 e t + c 2 t e t .

Then

y ( t ) = c 1 e t + c 2 ( t e t + e t ) .

When t = 0 , we have y ( 0 ) = c 1 and y ( 0 ) = c 1 + c 2 . Applying the initial conditions, we obtain

c 1 = 1 c 1 + c 2 = 0 .

Thus, c 1 = 1 , c 2 = 1 , and the solution to the initial value problem is

y ( t ) = e t + t e t .

This solution is represented in the following graph. At time t = 2 , the mass is at position y ( 2 ) = e −2 + 2 e −2 = 3 e −2 0.406 m below equilibrium.

This figure is the graph of y(t) = e^−t + te^−t. The horizontal axis is labeled with t and is scaled in increments of even tenths. The y axis is scaled in increments of 0.5. The graph passes through positive one and decreases with a horizontal asymptote of the positive t axis.

To calculate the velocity at time t = 1 , we need to find the derivative. We have y ( t ) = e t + t e t , so

y ( t ) = e t + e t t e t = t e t .

Then y ( 1 ) = e −1 0.3679 . At time t = 1 , the mass is moving upward at 0.3679 m/sec.

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Suppose the following initial-value problem models the position (in feet) of a mass in a spring-mass system at any given time. Solve the initial-value problem and graph the solution. What is the position of the mass at time t = 0.3 sec? How fast is it moving at time t = 0.1 sec? In what direction?

y + 14 y + 49 y = 0 , y ( 0 ) = 0 , y ( 0 ) = 1

y ( t ) = t e −7 t
This figure is the graph of y(t) = te^−7t. The horizontal axis is labeled with t and is scaled in increments of tenths. The y axis is scaled in increments of 0.5. The graph passes through the origin and has a horizontal asymptote of the positive t axis.
At time t = 0.3 , y ( 0.3 ) = 0.3 e ( −7 * 0.3 ) = 0.3 e −2.1 0.0367 . The mass is 0.0367 ft below equilibrium. At time t = 0.1 , y ( 0.1 ) = 0.3 e −0.7 0.1490 . The mass is moving downward at a speed of 0.1490 ft/sec.

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Solving a boundary-value problem

In [link] f. we solved the differential equation y + 16 y = 0 and found the general solution to be y ( t ) = c 1 cos 4 t + c 2 sin 4 t . If possible, solve the boundary-value problem if the boundary conditions are the following:

  1. y ( 0 ) = 0 , y ( π 4 ) = 0
  2. y ( 0 ) = 1 , y ( π 8 ) = 0
  3. y ( π 8 ) = 0 , y ( 3 π 8 ) = 2

We have

y ( x ) = c 1 cos 4 t + c 2 sin 4 t .
  1. Applying the first boundary condition given here, we get y ( 0 ) = c 1 = 0 . So the solution is of the form y ( t ) = c 2 sin 4 t . When we apply the second boundary condition, though, we get y ( π 4 ) = c 2 sin ( 4 ( π 4 ) ) = c 2 sin π = 0 for all values of c 2 . The boundary conditions are not sufficient to determine a value for c 2 , so this boundary-value problem has infinitely many solutions. Thus, y ( t ) = c 2 sin 4 t is a solution for any value of c 2 .
  2. Applying the first boundary condition given here, we get y ( 0 ) = c 1 = 1 . Applying the second boundary condition gives y ( π 8 ) = c 2 = 0 , so c 2 = 0 . In this case, we have a unique solution: y ( t ) = cos 4 t .
  3. Applying the first boundary condition given here, we get y ( π 8 ) = c 2 = 0 . However, applying the second boundary condition gives y ( 3 π 8 ) = c 2 = 2 , so c 2 = −2 . We cannot have c 2 = 0 = −2 , so this boundary value problem has no solution.
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Key concepts

  • Second-order differential equations can be classified as linear or nonlinear, homogeneous or nonhomogeneous.
  • To find a general solution for a homogeneous second-order differential equation, we must find two linearly independent solutions. If y 1 ( x ) and y 2 ( x ) are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by
    y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) .
  • To solve homogeneous second-order differential equations with constant coefficients, find the roots of the characteristic equation. The form of the general solution varies depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.
  • Initial conditions or boundary conditions can then be used to find the specific solution to a differential equation that satisfies those conditions, except when there is no solution or infinitely many solutions.

Key equations

  • Linear second-order differential equation
    a 2 ( x ) y + a 1 ( x ) y + a 0 ( x ) y = r ( x )
  • Second-order equation with constant coefficients
    a y + b y + c y = 0

Classify each of the following equations as linear or nonlinear. If the equation is linear, determine whether it is homogeneous or nonhomogeneous.

x 3 y + ( x 1 ) y 8 y = 0

linear, homogenous

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( 1 + y 2 ) y + x y 3 y = cos x

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x y + e y y = x

nonlinear

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y + 4 x y 8 x y = 5 x 2 + 1

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y + ( sin x ) y x y = 4 y

linear, homogeneous

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y + ( x + 3 y ) y = 0

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For each of the following problems, verify that the given function is a solution to the differential equation. Use a graphing utility to graph the particular solutions for several values of c 1 and c 2 . What do the solutions have in common?

[T] y + 2 y 3 y = 0 ; y ( x ) = c 1 e x + c 2 e −3 x

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[T] x 2 y 2 y 3 x 2 + 1 = 0 ; y ( x ) = c 1 x 2 + c 2 x −1 + x 2 ln ( x ) + 1 2

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[T] y + 14 y + 49 y = 0 ; y ( x ) = c 1 e −7 x + c 2 x e −7 x

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[T] 6 y 49 y + 8 y = 0 ; y ( x ) = c 1 e x / 6 + c 2 e 8 x

Find the general solution to the linear differential equation.

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y 3 y 10 y = 0

y = c 1 e 5 x + c 2 e −2 x

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y 7 y + 12 y = 0

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y + 4 y + 4 y = 0

y = c 1 e −2 x + c 2 x e −2 x

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4 y 12 y + 9 y = 0

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2 y 3 y 5 y = 0

y = c 1 e 5 x / 2 + c 2 e x

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3 y 14 y + 8 y = 0

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y + y + y = 0

y = e x / 2 ( c 1 cos 3 x 2 + c 2 sin 3 x 2 )

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5 y + 2 y + 4 y = 0

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y 121 y = 0

y = c 1 e −11 x + c 2 e 11 x

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8 y + 14 y 15 y = 0

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y + 81 y = 0

y = c 1 cos 9 x + c 2 sin 9 x

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y y + 11 y = 0

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2 y = 0

y = c 1 + c 2 x

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y 6 y + 9 y = 0

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3 y 2 y 7 y = 0

y = c 1 e ( ( 1 + 22 ) / 3 ) x + c 2 e ( ( 1 22 ) / 3 ) x

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4 y 10 y = 0

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36 d 2 y d x 2 + 12 d y d x + y = 0

y = c 1 e x / 6 + c 2 x e x / 6

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25 d 2 y d x 2 80 d y d x + 64 y = 0

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d 2 y d x 2 9 d y d x = 0

y = c 1 + c 2 e 9 x

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Solve the initial-value problem.

y + 5 y + 6 y = 0 , y ( 0 ) = 0 , y ( 0 ) = −2

y = −2 e −2 x + 2 e −3 x

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y + 2 y 8 y = 0 , y ( 0 ) = 5 , y ( 0 ) = 4

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y + 4 y = 0 , y ( 0 ) = 3 , y ( 0 ) = 10

y = 3 cos ( 2 x ) + 5 sin ( 2 x )

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y 18 y + 81 y = 0 , y ( 0 ) = 1 , y ( 0 ) = 5

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y y 30 y = 0 , y ( 0 ) = 1 , y ( 0 ) = −16

y = e 6 x + 2 e −5 x

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4 y + 4 y 8 y = 0 , y ( 0 ) = 2 , y ( 0 ) = 1

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25 y + 10 y + y = 0 , y ( 0 ) = 2 , y ( 0 ) = 1

y = 2 e x / 5 + 7 5 x e x / 5

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y + y = 0 , y ( π ) = 1 , y ( π ) = −5

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Solve the boundary-value problem, if possible.

y + y 42 y = 0 , y ( 0 ) = 0 , y ( 1 ) = 2

y = ( 2 e 6 e −7 ) e 6 x ( 2 e 6 e −7 ) e −7 x

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9 y + y = 0 , y ( 3 π 2 ) = 6 , y ( 0 ) = −8

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y + 10 y + 34 y = 0 , y ( 0 ) = 6 , y ( π ) = 2

No solutions exist.

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y + 7 y 60 y = 0 , y ( 0 ) = 4 , y ( 2 ) = 0

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y 4 y + 4 y = 0 , y ( 0 ) = 2 , y ( 1 ) = −1

y = 2 e 2 x 2 e 2 + 1 e 2 x e 2 x

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y 5 y = 0 , y ( 0 ) = 3 , y ( −1 ) = 2

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y + 9 y = 0 , y ( 0 ) = 4 , y ( π 3 ) = −4

y = 4 cos 3 x + c 2 sin 3 x , infinitely many solutions

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4 y + 25 y = 0 , y ( 0 ) = 2 , y ( 2 π ) = −2

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Find a differential equation with a general solution that is y = c 1 e x / 5 + c 2 e −4 x .

5 y + 19 y 4 y = 0

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Find a differential equation with a general solution that is y = c 1 e x + c 2 e −4 x / 3 .

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For each of the following differential equations:

  1. Solve the initial value problem.
  2. [T] Use a graphing utility to graph the particular solution.

y + 64 y = 0 ; y ( 0 ) = 3 , y ( 0 ) = 16

a. y = 3 cos ( 8 x ) + 2 sin ( 8 x )
b.
This figure is a periodic graph. It has an amplitude of 3.5. Both the x and y axes are scaled in increments of 1.

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y 2 y + 10 y = 0 y ( 0 ) = 1 , y ( 0 ) = 13

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y + 5 y + 15 y = 0 y ( 0 ) = −2 , y ( 0 ) = 7

a. y = e ( −5 / 2 ) x [ −2 cos ( 35 2 x ) + 4 35 35 sin ( 35 2 x ) ]
b.
This figure is a graph of an oscillating function. The x and y axes are scaled in increments of even numbers. The amplitude of the graph is decreasing as x increases.

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(Principle of superposition) Prove that if y 1 ( x ) and y 2 ( x ) are solutions to a linear homogeneous differential equation, y + p ( x ) y + q ( x ) y = 0 , then the function y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) , where c 1 and c 2 are constants, is also a solution.

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Prove that if a, b, and c are positive constants, then all solutions to the second-order linear differential equation a y + b y + c y = 0 approach zero as x . ( Hint: Consider three cases: two distinct roots, repeated real roots, and complex conjugate roots.)

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Questions & Answers

Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
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Renato
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Stoney Reply
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Adin Reply
?
Kyle
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Adin
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biomolecules are e building blocks of every organics and inorganic materials.
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research.net
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sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
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for screen printed electrodes ?
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s. Reply
of graphene you mean?
Ebrahim
or in general
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in general
s.
Graphene has a hexagonal structure
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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