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Exercises 4.2 Reduction of Order
and
y"+ 9y = (sin 3x)un+ 6(cos 3z)v! = 0 or u" + 6(cot 3x)u' =
0.
If w = u' we obtain the linear first-order equation w1+ 6(cot
3x)w ~ 0 which has the integrati:..
factor e(i J cot'ixdx = sin2 3x. Now
-^-[(sin23x)w] 0 gives (sin23.t)m-! = c.
Therefore w = v! = ecsc2 3.x and u = c\ cot 3.x. A second
solution is y-2 = cot 3xsin 3a: = cos 3x.
5. Define y = u(x )cosh x so
y' = usinh x+ u'cosh x, y" = u"cosh x+ 2u'sinh x ucosh x
and
y" y = (cosh x)u" + 2(sinli x)u' = 0 or un + 2(tanh.r)u/0 .
If w = v! we obtain the linear first- order equation w' + 2(tanh
x)w = 0 w hich has the integrati:.-
factor e2f Tanhx dx = cosh2x. Now
[(cosh2x)w\ - 0 gives (cosh2x)w c.LJLJb
Therefore w == u- csech2x and u= c tanh x. A second solution is
t/2 tanh a; cosh x = sinh x.
6 . Define y = u(x )e0X so
y1 5e5xu + e5xuf, y" = e5xu" + 10e?xul + 25eaxu
and
y" - 25y = (?x{u"+ 10u') = 0 or u" + 10u' = 0 .
If w = v! we obtain the linear first- order equation w1+ 10; = 0
which has the integ rating fac:e i o f d x = e i 0x_ N o w
4~ [e10;rU] = 0 gives e10xw = c.dx
Therefore w = uf = a?-10* and u = cie- 10x. A second solution is
y2 e- ^x(^x ( ~ox,
7. Define y = u(x )e2x^ so
y - e2x/3u e2x'' ur. y" = + -e2xi'Aii' -1- | e2x'/3uO O i/
and
9y " - Y ly + Ay = 9e2:c/V = 0.
Therefore u" = 0 and u = cyx + c-2 - Taking c\ = 1 and C2 = 0 we
see tha t a second solution
ij2 = x c2*'1 .
8 . Define y ~ -u(x)ex^ so
y ^e x^ u + ex^u ' < y " = + ^ex^'iul + e :S/,5w
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Exercises 4.2 Reduction of Order
A second solution is ij2= x2sin (In a;).
15. Identifying P(x) - 2(1 + x) j ( l - 2a; - x 2) we have
e - / '2(1-Hr)d x /{ l2xx2) ( } n { \ - 2 X - . r 2 )1)2= {X+ 1
)j --------------, ---------------dx = (X + 1 )j ------- dx
= ( . + i ) /
(x + 1 ) 2
1 2a x2
( * + l )2
2 1
(a: + 1)>
dx = (x + 1) J{x+ iy
- 1 dx
. x + 1 X = - 2 - r- - x.
In1 + X
1 X
(x+ 1)
A second s olution is 1/2 = x 2 + x + 2.
16. Identifying P(x) 2x j ( l x 2} we have
y 2 = I e ~ f -2xdx/(i- a?)dx = j e - H '- ^ d x = j Y ^ 2 d x =
1
A second s olution is y2 = ln | (l + x ) /( l x)\ .
17. Define y = u{x )e~ 2x so
y' - - 2ue_2x + u'e~'2x, y" = i"e ~ 2x - 4u'e~'2x + 4ue~2x
and
y" - 4y = e~2xu" - 4e 2:V = 0 or u" - 4v! = 0.
If w = it! wo obta in the linear first-order equation w' 4w = 0
which has the integrating
e~4f dx = e~4x. Now
-~[e~4xw] = 0 gives e~4xw = e.dx
Therefore w v! ce4x and u cieix . A second solution is y2
e~2xe4x = e2x. We
observation that a particular solution is yp 1/2. The general
solution is
2x , 2r 1y = cl e + c2e
18. Define y = u(x) 1 so
y' = u', y" = v " and y" + t j = u"+ u = 1.
If w = u' we obta in the linear first- order equation w' + w = 1
which has the integrating
eJ dx = ex. Nowd_
dx[exw] = ex g i v e s exw = ex + c.
Therefore w = u' = 1 + ce x and u = x + e.\e x+ r;2 . The
general solution is
y = u = x+ c\ e~x + c2.
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Exercises 4.2 Reduction of Order
Define y = .(a:)eT so
y = u,(f -i-iiex. y" = u"ex + 2u'ex + uex
and
y" - 3y'+ 2y = exu" - exu = 5e3*
If ; = u' we obta in the linear first- order equation w1 w =
5e2;r which has the integrating factor
r~ J dx = e~x. Now
= oex gives e~xw ~ 5e'1 + ci.dx
Therefore w u' 5e2x + C]_ex and u = | e2a: + cie ,T-rC2- The
general solution is
y = uex = ^e3x + cie 2u: -i- c2ex.z
Define y = u(x )ex so
y> = uer + u'ex , y" = u"e*+ 2u'ex + ue*
r.l'A
y" 4y1+ 3y = e1:u" exu' = x.
I: u: = v! we obtain the linear first- order equation wf 2w ~
xe~x which has the integrating factorJ 2 , l x = e - 2x_ N q w
dr_0 ,. , _
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Exercises 4.2 Reduction of Order
p b x /a
- (, m i x j e - { b l a^ r 2 m \ ) x ^ x
1 e m , x e - ( b / a ^ 2 m i ) x (mi - b/2a)
and let. y i = emyX be a solution. Then a second solution is
given by
b/a + 2mi
^ (b/am i )x
b/a+ 2m\
T hus, w hen m-j ^ b/2a, a second solution is given by 1 / 2 -
em'2'x where m 2 = b/a
When m-i b/2a a second solution is given by
U2 = em'lXJdx - x enilX.
(c) T he functions
sin a; = ~ (e v:r e iX) cos = ^(ew + e lx)/ z
sinh 2; = e_;i;) cosh./: = ^ (ex + e~x)
are all expressible in terms of exponential functions.
22. We have y[ 1 and y'{ = 0, so xy'[ xy\ -fy i = 0 x+ x 0 and y
i(x ) x is a. solution of :
differential equation. Letting y = u(x )y i(x ) = xu(x) we
get
y1= x n!(x ) + u(x) and y" = x u"{x ) + 2u'{x ).
Then x y" xy' + y = x 2u" + 2xu' x2v? x u+ x u = x2u" (x2 2x)u'
= 0 . If we make *.
substitution w = -u'. the linear first- order differential
equation becomes x2wf(x2x)w = 0 . wh:
is separable:
dw /. 1 ' | U'
- = ( l )dxx x J
dx v x '
dw
w
hiw = x In x + c
exW C] .
xThen v' - ciex jx and u - c\ j exdx /x . To integrate ex/x we
use the series representation for
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