Differential Equations_ Variation of Parameters

Diff Differ eren enti tial al Equat Equatio ions: ns: Varia ariati tion on of Para Parame mete terrs

http: http:// //ww www w.cli .cliff ffsn snot otes es.c .com om/s /stu tudy dy_g _gui uide de/V /Var aria iati tionon-of of-P -Par aram amet eter ers. s.top topic icA. A... ..

CliffsNotes

Home > Math > Differential Equations > Variation Variation of Parameters

Review and Introduction

Variation of Parameters

Differentiation Partial Differentiation

For the differential equation

Integration Techniques o f Indefinite Integration Partial Integration Introduction to Differential Equations First-Order Equations

the method of undetermined coefficients works only when the coefficients a, b, and c are are constants and the right-hand term d ( x ) is of a s pecial form. If these restrictions do not apply to a given nonhomogeneous linear differential differential equation, then a more powerful method o f determining a particular solution is needed: the method known as variation of parameters. parameters. The first step is to obtain the general solution of the corresponding homogeneous equation, which will have the form

Exact Equations Integrating Factors Separable Equations First-Order Homogeneous Equations

where y and y are known functions. The next step is to vary the parameters; parameters; that 1

2

is, to replace the constants c and c by (as yet unknown) functions v ( x ) and v ( 1

2

1

2

x ) to obtain the form of a particular solution y of the given nonhomogeneous equation:

First-Order Linear Equations Bernoulli's Equation Second-Order Equations

Linear Combinations, Linear Independence Second-Order Linear Equations

The goal is to determine these functions v and v . Then, since the functions y and 1

2

1

y are already known, the expression above for y yields a particular solution of the 2

nonhomogeneous equation. Combining y with y then gives the general solution of h

the non-homogeneous differential equation, as guaranteed by Theorem B. Since there are two unknowns to be determined, v and v , two equations or 1

2

conditions are required to obtain a solution. One of these conditions will naturally be satisfying the given differential equation. But another condition will be i mposed first. Since y will be substituted into equation (*), its derivatives must b e evaluated. The first derivative of y is

Second-Order Homogeneous Equations Constant Coefficients

Now, to simplify the rest of the process—and to produce the first condition on v

1

and v —set 2

The Method of Undetermined Coefficients Variation of Parameters Cauchy -Euler Equidimensional

1 f6

This will always be the first condition in determining v and v ; the second condition 1

2

will be the satisfaction of the given differential equation (*). equation (*).

26/03/2013 07 47

Differential Equations: Variation of Parameters

Equation Reduction of Order Power Series Introduction to Power Series

http://www.cliffsnotes.com/study_guide/Variation-of-Parameters.topicA...

Example 1: Give the general solution of the differential equation y ″ + y = tan x . Since the nonhomogeneous right-hand term, d = tan x , is not of the special form the method of undetermined coefficients can handle, variation of parameters is required. The first step is to obtain the g eneral solution of the corresponding homogeneous equation, y ″ + y = 0. The auxiliary polynomial equation is

Taylor Series Solutions of Differential Equations The Laplace Transform

whose roots are the distinct conjugate complex numbers m = ± i = 0 ± 1 i . The general solution of the homogeneous equation is therefore

Linear Transformations The Laplace Transform Operator

Now, vary the parameters c and c to obtain 1

2

Solving Differential Equations Ap pl yi ng Differential Equations

Differentialtion yields

Ap pl ic ati on s o f First -Order Equations Ap pl ic ati on s o f Second -Order Equations

Nest, remember the first condition to be imposed on v and v : 1

2

Related Topics: Calculus

that is,

Precalculus

This reduces the expression for y′ to

so, then,

Substitution into the given nonhomogeneous equation y ″ + y = tan x yields

Therefore, the two conditions on v and v are 1

2 f6

2

26/03/2013 07 47



To solve these two equations for v ′ and v ′, first multiply the first equation by sin 1

2

x ; then multiply the second equation by cos x :

Adding these equations yields

Substituting v ′ = sin x back into equation (1) [or equation (2)] then gives 1

Now, integrate to find v and v (and ignore the constant of integration in each 1

2

case):

and

Therefore, a particular solution of the given nonhomogeneous differential equation is

Combining this with the general solution of the corresponding homogeneous equation gives the general solution of the nonhomogeneous equation:

In general, when the method of variation of parameters is applied to the second-order nonhomogeneous linear differential equation

3 f6

26/03/2013 07 47



with y = v ( x ) y + v ( x ) y (where y = c y + c y is the general solution of 1

1

2

2

h

1

1

2

2

the corresponding homogeneous equation), the two conditions on v and v will 1

2

always be

So after obtaining the general solution of t he corresponding homogeneous equation ( y = c y + c y ) and varying the parameters by writing y = v y + v y , go h

1

1

2

2

1

1

2

2

directly to equations (1) and (2) above and solve for v ′ and v ′. 1

2

Example 2: Give the general solution of the differential equation

Because of the In x term, the right-hand side is not one of the special forms that the method of undetermined coefficients can handle; variation of parameters is required. The first step requires obtaining the g eneral solution of the corresponding homogeneous equation, y ″ – 2 y ′ + y = 0:

Varying the parameters gives the particular solution

and the system of equations (1) and (2) becomes

x

Cancel out the common factor of e

in both equations; then subtract the resulting

equations to obtain

Substituting this back into either equation (1) or (2) determines

Now, integrate (by parts, in both these cases) to obtain v and v from v ′ and v ′: 1

4 f6

2

2

2

26/03/2013 07 47



Therefore, a particular solution is

Consequently, the general solution o f the given nonhomogeneous equation is

Example 3: Give the general solution of the following differential equation, given 3

that y = x and y = x are solutions of its c orresponding homogeneous equation: 1

2

3

Since the functions y = x and y = x are linearly independent, Theorem A says 1

2

that the general solution of the corresponding homogeneous equation is

Varying the parameters c and c gives the form of a particular solution of the given 1

2

nonhomogeneous equation:

where the functions v and v are as yet undetermined. The two conditions on v 1

2

1

and v which follow from the method of variation of parameters are 2

3

2

4

which in this case ( y = x, y = x , a = x , d = 12 x ) become 1

2

Solving this system for v ′ and v ′ yields 1

2

from which follow

5 f6

26/03/2013 07 47



Therefore, the particular solution obtained is

and the general solution of the given nonhomogeneous equation is

Cliff's Notes Do you know where I can find [pre-calculus h omework] help on the weekends or whenever? What are limits in calculus? More Study Help

More Study Help

Connect with CliffsNotes

Shakespeare Central

CliffsNotes Films

Ab out Clif fs Notes

Unsubscribe from Our Newsletters

Test Prep & Cram Plans

Vocabulary Help: The Defining Twilight Series

Contact Us

CliffsNotes on Facebook

Ad vert ise w it h Us

CliffsNotes on Twitter

Teacher Resources

CliffsNotes on YouTube

Buy CliffsNotes Books and E-books

Download CliffsNotes Apps

Cliff's Notes Manga Editions

College Study Break

||

6 f6

26/03/2013 07 47

Differential Equations_ Variation of Parameters

Recommend Documents