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Ejercicios de matemtica III - 2
3a) y 5y + 6y = 2senx + 8
Si: !r)= r3 5r2+ 6r= "
r!r2
- 5r + 6)= "
r!r-2)!r-3) = " # entonces r$= " % r2= 2 % r3= 3
y&= c$ + c2e2x+ c3e3x
y'= ($ + (2e2x+ (3e3x
u1'+u2
'e
2x+u 3'e
3x=02 e
3 e
( 3x)=04 e
9 e
( 2x)+u3
' ( 3x)=2 senx+8
( 2x)+u3
' 0+u2
'
0+u2
'
u1
'=| 0 e
2xe
3x
0 2x2x
3x3 x
2 senx+8 4 x2 x 9x3x||1 e2x
e3x
0 2x2x
3x3x
0 4x2 x
9x3x|
=(2 senx+8)e5x
18 e5x12 e5x
=(2 senx+8)e5 x
6 e5x
=2 senx+8
6=
senx+43
u1
'=senx+4
3 interandou1' du=
senx+43
dx=1
3(cosx+4x)
u1=cosx
3+ 4
3x
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u2
'=|1 0 e
3x
0 0 3x3 x
0 2 senx+8 9x3 x|
|1 e
2xe
3x
0 2x2x
3x3 x
0 4x2 x
9x3 x
|
=(2 senx+8)3 e3x
18 e5x12 e5 x
=(2 senx+8)3 e3 x
6 e5x
=(2 senx+8)3 e2x
6=(senx+4)
u2'=(senx+4)e2x interando u2
'du=(senx+4)e2x dx
u2' du= ( senx ) e2xdx(4 e2x)dx=e2x(2 senxcox )
54
(e2x)2
u2=e2x(2
senx+cox )5 +2 e2x
u3
'=|1 e
2x0
0 2x2x
0
0 4x2x
2 senx+8|
|1 e2x
e3x
0 2x2x
3x3x
0 4x2x
9x3x|
=(2 senx+8)2 e2x
18 e5x12e5x
=(2 senx+8)2e2x
6 e5x
=(2 senx+8)2 e3x
6=(2 senx+8)e3x
3
u3
'=2(senx+4)e3x
3 interando
u3' du=2(senx+4 )e3x
3dx=
2
3(senx+4 )e3x dx
u3' du=2
3(senx)e3x dx+ 2
3(4 )e3x dx=2
3 (e3x(3 senxcosx )10 )+83 (e3x)
3
u3=e3x
(3 senx+cox )15
8
9e3x
y'= ($ + (2e2x+ (3e3x
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yp=cosx
3+4
3x+(e2x(2 senx+cox )5 +2e2x)e2x+(e3x(3 senx+cox )15 89
e3x)e3x
yp=cosx
3
+4
3
x+
(( 2 senx+cox )
5
+2
)+(
( 3 senx+cox )
15
8
9
)
yg=c1+c
2e2x+c
3e3x
cosx
3+4
3x+((2 senx+cox )5 +2)+((3 senx+cox )15 89 )
3*)y + y = senxcosx
Si: !r)= r3+ r$= "
r!r2+ )= "
r!r-2i)!r+2i) = " # entonces r$= " % r2= 2i % r3= -2i
y&= c$ + c2cos2x + c3sen2x
y'= ($ + (2cos2x + (3sen2x
{
u1
'+u2
'cos2x+u
3
'sen 2x=0
0+u2
' (2 sen 2x )+u3
' (2cos2x )=0
0
+u
2
'
(4 c os 2x
)+u
3
'
(4 sen 2x
)=senxcosx
u1'=
| 0 cos 2x sen 2x
0 2 sen 2x 2cos2xsenxcosx 4cos2x 4 sen 2x|
|1 cos 2x sen 2x0 2 sen 2x 2cos2x0 4cos2x 4 sen 2x|
=senxcosxcos2x 2cos2x
8 [( sen 2x )2+(cos2x )2]=
2 senxcosx(cos2x)2
8 =
sen 2x (co8
Si : r=2x entonces dr=2dx
u1' du=sen 2x (cos 2x )2 dx
8= senr(cosr)
2
dr
16=
1
16 [senr( senr )3 ] dr= 1
16[
( senr )2
2(senr)4
4]
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u1= 1
16 [( senr )2
2
(senr ) 4
4 ]= 116[(sen 2x )2
2
( sen 2x )4
4 ]
(sen 2x )2+(cos2x)2 }
8
u2
'=|1 0 sen 2x0 0 2cos 2x
0 sexcosx 4 sen 2x||
1 cos 2x sen 2x
0 2 sen 2x 2cos 2x0 4cos2x 4 sen 2x|
=senxcosx(2cos2x)
Si : r=2x entonces dr=2dx
(senrco sr
16)dr
u2' du=sen 2xcos 2x
8dx==1
32(senr)2 =
132
(sen 2x )2
u2=132
(sen 2x )2
(sen 2x)
8 [2+(cos2x)2]= (sen 2x ) sen 2 x
8=(
sen 4 x )16
u3
'=|1 cos 2x 00 2 sen 2x 0
0 4 cos 2x senxcosx||
1 cos 2x sen 2x
0 2 sen 2x 2cos2x0 4cos 2x 4 sen 2x|
=(2 sen 2x )(senxcosx)
Si : r=x entonces dr=dx
u3' du=( sen 4x )
16dx=( senr64)dr=1128 (senr )2=1128 (sen 4x)2
u3=1128
(sen 4x)2
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y'= ($ + (2cos2x + (3sen2x
y'=1
16 [(sen 2x )2
2
( sen 2x )4
4 ] + 132 ( sen 2x )2 cos2x 1128 (sen 4x )2 sen2x
y= c$ + c2cos2x + c3sen2x+1
16 [(sen 2x )2
2
( sen 2x )4
4 ] +132
( sen 2x )2 cos2x 1
128(sen 4x )2
sen2x
