variation of parameters a lecture in engiana. background the procedure used to find a particular...

Variation of Parameters

A Lecture in ENGIANA

Background• The procedure used to find a particular solution

yp of a linear first-order differential equation on

an interval is applicable to linear higher-order DEs as well.

• To adapt the method of variation of parameters to a linear second-order DE

we rewrite the equation into standard form:

)x(gy)x(a'y)x(a''y)x(a 012

)x(fy)x(Q'y)x(P''y

Assumptions• We seek a solution of the form

where y1 and y2 form a fundamental set of

solutions on an interval of the associated homogeneous form.

• Differentiating twice, we get

)x(y)x(u)x(y)x(uy 2211p

'22

'22

'11

'11

'p uyyuuyyuy

'2

'2

''22

'2

'2

''22

'1

'1

''11

'1

'1

''11

''p yuuyuyyuyuuyuyyuy

Assumptions

Substituting yp and its derivatives into the standard form of a second order DE and grouping terms yields

'2

'2

'1

'1

'22

'11

'2

'2

''22

'1

'1

''11

2'2

''221

'1

''11

p'p

''p

uyuy

uyuyPyuuyyuuy

QyPyyuQyPyyu

y)x(Qy)x(Py

Assumptions

'2

'2

'1

'1

'22

'11

'22

'11

21

'2

'2

'1

'1

'22

'11

'2

'2

''22

'1

'1

''11

2'2

''221

'1

''11

uyuy

uyuyPuydx

duy

dx

d

0u0u

uyuy

]uyuy[Pyuuyyuuy

]QyPyy[u]QyPyy[u

Assumptions

'

2'2

'1

'1

'22

'11

'22

'11

'2

'2

'1

'1

'22

'11

'22

'11

uyuyuyuyPuyuydx

d

uyuyuyuyPuydx

duy

dx

d

Because we seek to determine two unknown functions u1 and u2, we need two independent equations. We can obtain these equations by making the assumption that y1u1’ + y2u2’ = 0.

Assumptions

)x(fuyuy

)x(fy)x(Qy)x(Py

,Hence

uyuy0P0dx

d

uyuyuyuyPuyuydx

d

y)x(Qy)x(Py

'2

'2

'1

'1

p'p

''p

'2

'2

'1

'1

'2

'2

'1

'1

'22

'11

'22

'11

p'p

''p

)x(fy

0yW

y)x(f

y0W

yy

yyW

whereW

Wu

W

Wu

is

)x(fu'yu'y

0uyuy

systemtheofsolutionthe,Rules'CramerBy

'1

12'

2

21'

2'1

21

2'2

1'1

'22

'11

'22

'11

Assumptions

• The functions u1 and u2 are found by integrating u1’ and u2’, respectively.

• The determinant W is the Wronskian of y1 and y2.

• By linear independence of y1 and y2 on an interval, we know that W(y1(x), y2(x)) 0 for every x in the interval.

Summary of the Method

• Usually, it is not a good idea to memorize formulas in lieu of understanding a procedure.

• However, the foregoing procedure is too long and complicated to use each time we wish to solve a differential equation.

• In this case, it is more efficient to simply use formulas.


Step 1. Put a2y’’ + a1y’ + a0y = g(x) into

the standard form

y’’ + Py’ + Qy = f(x).

Step 2. Find the complementary function yc = c1y1 + c2y2.

Step 3. Compute the Wronksian

W(y1(x), y2(x)).

Step 4. Compute for W1 and W2.


Step 5. Solve for u1 and u2 by integrating u1’ = W1/W and u2’ = W2/W.

Step 6. Express the particular solution as

yp = u1y1 + u2y2

Step 7. The general solution of the equation is then

y = yc + yp.

Example

Find the general solution of

(D2 + 1)y = cscx.

xsinuxcosuy

Let

:yfind,Next

xsincxcoscy

,So

im

01m

:firstyFind

xcscy)1D(

:Solution

21p

p

21c

2

c

2

1)xsin)(x(sin)x(cos

xcosxsin

xsinxcosW

:Wfind,Next

xcsc)x(cosu)xsin(u

0xsinuxcosu

xcsc)x(sindx

du)x(cos

dx

du

0xsinuxcosu

havewe,sderivationpreviousFrom

xsinuxcosuy

2

'2

'1

'2

'1

'2

'1

'2

'1

21p

xcot0)x)(cscx(cosxcscxsin

0xcosW

1)x)(cscx(sin0xcosxcsc

xsin0W

:WandWfind,Then

2

1

21

xsinlndx)x(cotu

xdx)1(u

:uanduforsolve,Then

xcot1

xcot

W

Wu

11

1

W

Wu

uanduforsolve,Then

2

1

21

2'2

1'1

'2

'1

xsinlnxsinxcosxxsincxcoscy

yyy

,So

xsin)xsin(lnxcosxy

xsincxcoscy

xsinuxcosuy

xsincxcoscy

,Thus

21

pc

p

21c

21p

21c

Note on constants of integration

When computing the indefinite integrals of u1’ and u2’, we need not introduce any constants. This is because

y = yc + yp

= c1y1 + c2y2 + (u1 + a1)y1 + (u2 + a2)y2

= (c1 + a1)y1 + (c2 + a2)y2 + u1y1 + u2y2

= C1y1 + C2y2 + u1y1 + u2y2

Example


y’’ + 2y’ + y = e-tlnt.

t2

t1c

2

2

c

t

tececy

,Hence

0)1m(

01m2m

:yFind

tlney'y2''y

:Solution

t2

ttttt

ttt

tt

ttt'2

t'1

t'2

t'1

t2

t1p

p

eW

)e)(te()tee)(e(W

teee

teeW

,Then

tlne)tee(u)e(u

0teueu

teueuy

:yFind

tlneW

0)tlne)(e(W

tlnee

0eW

and

tlnteW

)tlne)(te(0W

teetlne

te0W

t22

tt2

tt

t

2

t21

tt1

ttt

t

1

ttlnttdtlnu

t4

1tlnt

2

1tdtlntu

:uanduforsolve,Then

tlne

tlne

W

Wu

tlnte

tlnte

W

Wu

uanduforsolve,Then

2

221

21

t2

t22'

2

t2

t21'

1

'2

'1

pc

tt22p

t2

t1c

t2

t1p

t2

t1c

yyy

,So

te)ttlnt(e)t4

1tlnt

2

1(y

tececy

teueuy

tececy

,Thus

t2t2t2

t1

t2t2t2t2t2

t1

tt22t2

t1

et4

3tlnet

2

1tececy

ettlnetet4

1tlnet

2

1tececy

te)ttlnt(e)t4

1tlnt

2

1(tececy

Higher-Order Equations

The variation of parameters method for non-homogeneous second-order differential equations can be generalized to linear nth-order equations that have been put into the standard form

)x(fy)x(P'y)x(P...y)x(Py 01)1n(

1n)n(

)x(fuy...uyuy

0uy...uyuy

0uy...uyuy

equationsnthebysolvedaren...,,2,1k,uwhere

)x(y)x(u...)x(y)x(u)x(y)x(uy

issolutionparticularathen

,functionarycomplement

theisyc...ycycyIf

'n

)1n(n

'2

)1n(2

'1

)1n(1

'n

'n

'2

'2

'1

'1

'nn

'22

'11

'k

nn2211p

nn2211c

''3

''2

''1

'3

'2

'1

321

''2

''1

'2

'1

21

3''3

''1

'3

'1

31

2''3

''2

'3

'2

32

1

3'3

2'2

1'1

k'k

yyy

yyy

yyy

W

and

)x(fyy

0yy

0yy

W,

y)x(fy

y0y

y0y

W,

yy)x(f

yy0

yy0

W

,W

Wu,

W

Wu,

W

Wu

,systemorderrd3aFor

n...,,2,1kwhereW

Wu

givesthenRules'Cramer

Example


y’’’ + y’ = tanx

xsincxcosccy

,So

i,0m

0)1m(m

0mm

:yGet

xtan'y'''y

:Solution

321c

2

3

c

1W

xcosxsinW

xsinxcos

xcosxsin1

xsinxcos0

xcosxsin0

xsinxcos1

W

,Then

xtan)xsin('u)xcos('u)0('u

0)x(cos'u)xsin('u)0('u

0)x(sin'u)x(cos'u)1('u

:fftheupsetwe,uand,u,uforsolveTo

xsinuxcosuuy

:isyofformthemeansThis

22

321

321

321

321

321p

p

xcos

xsinxtanxsin

xtanxcos

0xsin1

xtanxcos0

0xsin0

0xcos1

W

xsinxtanxcosxsinxtan

xcos01

xsinxtan0

xcos00

xsin01

W

xtanxcosxsin

xsinxcosxtan

xsinxcosxtan

xcosxsin0

xsinxcos0

W

2

3

2

1

xcosxsecxcos

xcos1

xcos

xsin

W

W'u

xsin1

xsin

W

W'u

xtan1

xtan

W

W'u

,Hence

223

3

22

11

Remarks

The variation of parameters method has a distinct advantage over the method of undetermined coefficients in that it will always yield a particular solution yp provided that the associated homogeneous equation can be solved.

Examples & Exercises

Find the general solution of the following:

1) (D2 + 1)y = sec3x

2) (D2 + 1)y = tanx

3) (D2 + 1)y = secxcscx

4) (D2 – 2D + 1)y = e2x(ex+1)-2

5) (D2 – 3D + 2)y = cos(e-x)

6) (D2 – 1)y = e-2xsin(e-x)7) y’’’ – y’ = x [use VOP and compare with MUC]

variation of parameters a lecture in engiana. background the procedure used to find a particular...

Documents