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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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The nanotechnology is as new science, to scale nanometric
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nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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Practice Key Terms 7

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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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