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Previous Work Variation of Parameters Conclusion

MATH 312

Section 4.6: Variation of Parameters

Prof. Jonathan Duncan

Walla Walla College

Spring Quarter, 2007

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Outline

1 Previous Work

2 Variation of Parameters

3 Conclusion

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Why we Need Another Method

We now have a procedure for solving some linear differentialequations with constant coefficients, but it is far from complete.

Example

The following differential equations can not be solved byannihilators and variation of parameter (why not?):

y + y = cos2 x

x2y + xy + x2 1

4 = x

34

2y + 2y + y = 4x

To solve such equations, we turn to the methods used in solving 1storder equations.

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Variation of Parameters with 1st Order DEs

When solving a first order non-homogeneous linear differentialequation, we used a method called variation of parameter to find aparticular solution yp.

Variation of Parameter

The first order linear differential equation dydx + P(x)y = f(x) hada general solution y = yc + yp where yc is the general solution tothe associated homogeneous equation. We found that:

yp = e

RP(x) dx

e

RP(x) dx

f(x) dx

Note:

The assumption with which we started is that yp(x) = u(x)yc(x).

How does this generalize to 2nd order DEs?

P i W k V i i f P C l i

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2nd Order Variation of Parameters

With second order differential equations, yc is a linear combinationof two linearly independent functions.

Variation of Parameter

The second order differential equationd2ydx2

+ P(x)dydx

+ Q(x)y = f(x) has a general solution y = yc + ypwhere yc = C1y1 + C2y2 is the solution to the associatedhomogeneous equation, and yp is a particular solution.

Note:

Since we now have two linearly independent solutions y1 and y2going into yc, we assume that yp(x) = u1(x)y1(x) + u2(x)y2(x) isa combination of both.

P i W k V i ti f P t C l i

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Variation of Parameters Formula

In order to make progress, we must differentiate yp and plug it intoy + P(x)y + Q(x)y = f(x).

yp = u1y1 + u2y2

yp = u1y

1 + u

1y1 + u2y

2 + u

2y2

yp = u1y

1 + 2u

1y

1 + u

1y1 + u2y

2 + 2u

2y

2 + u

2y2

ddx

u1y1 + u

2y2

+ P(u1y1 + u

2y2) + u

1y

1 + u

2y

2 = f(x)


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Continued Derivation. . .

We now choose u1 and u2 so that u

1y1 + u

2y2 = 0.

u1y1 + u

2y2 = 0

u

1y

1+ u

2y

2= f(x)

u1 =

0 y2f(x) y2

y1 y2y1 y2

u2 =

y1 0y1 f(x)

y1 y2y1 y2

u1 =y2f(x)W(y1, y2)

u2 =y1f(x)

W(y1, y2)


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2nd Order Examples

We now apply this formula to the examples we saw earlier.

Example

Solve the differential equation y + y = cos2 x.

Example

Solve the differential equation x2y + xy +x2 14y

= x

32 given

that y1 = x

12 cos x and y2 = x

12 sin x.

Example

Solve the differential equation 2y + 2y + y = 4x.


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Generalizing to Higher Order

This method works for second order linear differential equations,but can it be extended to higher order equations?

Variation of Parameters for Higher OrderIfy1 , y2 , . . . ,yn1 are a fundamental set of solutions to the homogeneous equation associated to

an(x)y(n) + an1(x)y(n1) + + a1(x)y + a0(x) = g(x)

Then, the particular solution is:yp = u1y1 + u2y2 + + unyn

where the uis are given by:

u

i

=

y1 y2 0 yny1 y

2 0 y

n

.

.

.

.

.

.. .

....

. ..

.

.

.

y(i1)1 y

(i1)2 f(x) y

(i1)n

.

.

.

.

.

.. .

....

. ..

.

.

.

y(n1)1 y

(n1)2 0 y

(n1)n

W(y1 , y2 , . . . , yn)


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Important Concepts

Things to Remember from Section 4.6

1 Using variation of parameters to solve linear differential

equations

2 The variation of parameter formula for second order lineardifferential equations

3 Generalization of variation of parameter to higher order lineardifferential equations

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