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Resuelva Por Coeficientes Indeterminados y Variacion de Parametros
Citation preview
Resuelva
xey '' 2y ' y
x
Solución
Aquí n 2 y x x
h 1 2y c e c xe por lo tanto
x x
p 1 2y v e v xe
Dado que x x
1 2y e ,y xe y x(x) e / x de la ecuación tenemos
1 1 2 2
1 1 2 2
x
1 2
xx x x
1 2
v ' y v ' y 0
v ' y ' v ' y ' (x)
v ' e v ' xe 0
ev ' e v ' e xe
x
Resolviendo este conjunto de ecuaciones, obtenemos 1 2v ' 1 y v' 1/ x . De
este modo,
1 1
2 2
v v ' dx 1dx x
1v v ' dx dx ln x
x
Sustituyendo estos valores obtenemos
x x
py xe xe ln x
Por lo tanto, la solución general es
x x x x
c p 1 2
x x x
1 3 3 2
y y y c e c xe xe xe ln x
y c e c xe xe ln x c c 1
y'' -2y'+y=e^x�x
Input:
y¢¢HxL - 2 y
¢HxL + yHxL �ãx
x
ODE classification:
second-order linear ordinary differential equation
Alternate forms:
y¢¢HxL + yHxL � 2 y
¢HxL +ãx
x
y¢¢HxL � 2 y
¢HxL - yHxL +ãx
x
Alternate form assuming x is positive:
ãx � x Iy¢¢HxL - 2 y
¢HxL + yHxLM
Differential equation solution: Approximate form Step-by-step solution
yHxL � c1 ãx
+ c2 ãx
x + ãx
x logHxLlogHxL is the natural logarithm»
Plots of sample individual solutions:
x
y
y
y¢ yH1L � 1
y¢H1L � 0
x
y
y
y¢ yH1L � 0
y¢H1L � 1
Generated by Wolfram|Alpha (www.wolframalpha.com) on March 23, 2014 from Champaign, IL.
© Wolfram Alpha LLC— A Wolfram Research Company1
Sample solution family:
1.5 2.0 2.5 3.0
x
10
20
30
40
50
60
y
Hsampling yH1L and y¢H1LL
Possible Lagrangian:
LIy¢, y, xM �
1
2
-ã-2 x
y2
+ ã-2 x
y¢2
+2 ã-x
y
x
y'' -2y'+y=e^x/x
Generated by Wolfram|Alpha (www.wolframalpha.com) on March 23, 2014 from Champaign, IL.
© Wolfram Alpha LLC— A Wolfram Research Company2
Solve -2â yHxL
âx+
â2yHxL
âx2
+ yHxL � ãx
x:
The general solution will be the sum of the complementary solution and
particular solution.
Find the complementary solution by solvingâ2
yHxLâx
2- 2
â yHxLâx
+ yHxL � 0:
Assume a solution will be proportional to ãΛ xfor some constant Λ.
Substitute yHxL � ãΛ xinto the differential equation:
â2
â x2
IãΛ xM - 2
â
â x
IãΛ xM + ã
Λ x � 0
Substituteâ2
âx2
IãΛ xM � Λ2 ãΛ xand
â
âxIãΛ xM � Λ ãΛ x
:
Λ2
ãΛ x
- 2 Λ ãΛ x
+ ãΛ x � 0
Factor out ãΛ x:
IΛ2
- 2 Λ + 1M ãΛ x � 0
Since ãΛ x ¹ 0 for any finite Λ, the zeros must come from the polynomial:
Λ2
- 2 Λ + 1 � 0
Factor:
HΛ - 1L2 � 0
Solve for Λ:
Λ � 1 or Λ � 1
The multiplicity of the root Λ � 1 is 2 which gives y1HxL � c1 ãx, y2HxL � c2 ãx
x as
solutions, where c1 and c2 are arbitrary constants.
The general solution is the sum of the above solutions:
yHxL � y1HxL + y2HxL � c1 ãx
+ c2 ãx
x
Determine the particular solution toâ2
yHxLâx
2+ yHxL - 2
â yHxLâx
� ãx
xby variation of
parameters:
List the basis solutions in ycHxL:yb1
HxL � ãxand yb2
HxL � ãxx
Compute the Wronskian of yb1HxL and yb2
HxL:WHxL �
ãx ãxx
â
âxHãxL â
âxHãx
xL �ãx ãx
x
ãx ãx + ãxx
� ã2 x
Let f HxL � ãx
x:
Let v1HxL � -à f HxL yb2HxL
WHxL â x and v2HxL � à f HxL yb1HxL
WHxL â x:
The particular solution will be given by:
ypHxL � v1HxL yb1HxL + v2HxL yb2
HxL
Compute v1HxL:v1HxL � -à 1 â x � -x
Compute v2HxL:v2HxL � à 1
x
â x � logHxL
The particular solution is thus:
ypHxL � v1HxL yb1HxL + v2HxL yb2
HxL � -Hãx
xL + ãx
x logHxL
Simplify:
ypHxL � ãx
x HlogHxL - 1L
The general solution is given by:
yHxL � ycHxL + ypHxL � c1 ãx
+ c2 ãx
x + ãx
x HlogHxL - 1L
Simplify the arbitrary constants:
Answer:
yHxL � ãx
x logHxL + c1 ãx
+ c2 ãx
x
Resuelva t t
y '' 6y ' 25y 2sen cos2 2
Solución
La ecuación característica es
2 6 25 0
Usando la formula cuadrática encontramos que sus raíces son:
2
6 6 4 253 i4
2
Estas raíces son un par conjugado complejo, de modo que la solución general es:
3t 3t
1 2y c e cos4t c e sen4t
Aquí (t) tiene la forma siguiente, con la variable independiente t reemplazando
a 1 2
1x,k 2,k 1 y
2 1 2(x) k sen x k cos x donde 1 2k ,k y son
constantes conocidas. Se asume una solución de la forma
py Asen x Bcos x
Donde A y B son constantes a ser determinadas
Donde tenemos
p
p
p
t ty Asen Bcos
2 2
A t B ty ' cos sen
2 2 2 2
A t B ty '' sen cos
4 2 4 2
Sustituyendo los resultados en la ecuación diferencial obtenemos
A t B t A t B t t tsen cos 6 cos sen 25 Asen Bcos
4 2 4 2 2 2 2 2 2 2
t t2sen cos
2 2
O de manera equivalente
99 t 99 t t tA 3B sen 3A B cos 2sen cos
4 2 4 2 2 2
Igualando los coeficientes de los términos similares tenemos
99 99A 3B 2 : 3A B 1
4 4
Luego tenemos que A 56 / 663 y B 20 / 663 de modo que tenemos que
la solución particular es
p
56 t 20 ty sen cos
663 2 663 2
Y la solución general es
3t 3t
c p 1 2
56 t 20 ty y y c e cos4t c e sen4t sen cos
663 2 663 2
y'' -6y'+25y=2senHt�2L-cosHt�2L
Input:
y¢¢HtL - 6 y
¢HtL + 25 yHtL � 2 sin
t
2
- cos
t
2
ODE classification:
second-order linear ordinary differential equation
Alternate forms:
y¢¢HtL + 25 yHtL + cos
t
2
� 6 y¢HtL + 2 sin
t
2
y¢¢HtL � 6 y
¢HtL - 25 yHtL + 2 sin
t
2
- cos
t
2
y¢¢HtL - 6 y
¢HtL + 25 yHtL � -1
2
+ ä ã-
ä t
2 -1
2
+ ä ãä t
2
Differential equation solution: Approximate form Step-by-step solution
yHtL � c1 ã3 t
sinH4 tL + c2 ã3 t
cosH4 tL +56
663
sin
t
2
-20
663
cos
t
2
Plots of sample individual solutions:
t
y
y
y¢ yH0L � 1
y¢H0L � 0
t
y
y
y¢ yH0L � 0
y¢H0L � 1
Generated by Wolfram|Alpha (www.wolframalpha.com) on March 23, 2014 from Champaign, IL.
© Wolfram Alpha LLC— A Wolfram Research Company1
Sample solution family:
0.2 0.4 0.6 0.8 1.0 1.2
t
-20
20
40
60
80
100
y
Hsampling yH0L and y¢H0LL
Possible Lagrangian:
LIy¢, y, tM �
1
2
-25 ã-6 t
y2
+ ã-6 t
y¢2
+ 2 ã-6 t
y 2 sin
t
2
- cos
t
2
y'' -6y'+25y=2sen(t/2)-cos(t/2)
Generated by Wolfram|Alpha (www.wolframalpha.com) on March 23, 2014 from Champaign, IL.
© Wolfram Alpha LLC— A Wolfram Research Company2
Solve -6â yHtL
ât+
â2yHtL
ât2
+ 25 yHtL � 2 sinI t
2M - cosI t
2M :
The general solution will be the sum of the complementary solution and
particular solution.
Find the complementary solution by solvingâ2
yHtLât
2- 6
â yHtLât
+ 25 yHtL � 0:
Assume a solution will be proportional to ãΛ tfor some constant Λ.
Substitute yHtL � ãΛ tinto the differential equation:
â2
â t2
IãΛ tM - 6
â
â t
IãΛ tM + 25 ã
Λ t � 0
Substituteâ2
ât2
IãΛ tM � Λ2 ãΛ tand
â
âtIãΛ tM � Λ ãΛ t
:
Λ2
ãΛ t
- 6 Λ ãΛ t
+ 25 ãΛ t � 0
Factor out ãΛ t:
IΛ2
- 6 Λ + 25M ãΛ t � 0
Since ãΛ t ¹ 0 for any finite Λ, the zeros must come from the polynomial:
Λ2
- 6 Λ + 25 � 0
Solve for Λ:
Λ � 3 + 4 ä or Λ � 3 - 4 ä
The roots Λ � 3 ± 4 ä give y1HtL � c1 ãH3+4 äL t, y2HtL � c2 ãH3-4 äL t
as solutions, where
c1 and c2 are arbitrary constants.
The general solution is the sum of the above solutions:
yHtL � y1HtL + y2HtL � c1 ãH3+4 äL t
+ c2 ãH3-4 äL t
Apply Euler's identity ãΑ+ä Β � ãΑcosH ΒL + ä ãΑ
sinH ΒL:yHtL � c1 Iã
3 tcosH4 tL + ä ã
3 tsinH4 tLM + c2 Iã
3 tcosH4 tL - ä ã
3 tsinH4 tLM
Regroup terms:
yHtL � Hc1 + c2L ã3 t
cosH4 tL + ä Hc1 - c2L ã3 t
sinH4 tL
Redefine c1 + c2 as c1 and ä Hc1 - c2L as c2, since these are arbitrary constants:
yHtL � c1 ã3 t
cosH4 tL + c2 ã3 t
sinH4 tL
Determine the particular solution toâ2
yHtLât
2+ 25 yHtL - 6
â yHtLât
� 2 sinI t
2M - cosI t
2M by
variation of parameters:
List the basis solutions in ycHtL:yb1
HtL � ã3 tcosH4 tL and yb2
HtL � ã3 tsinH4 tL
Compute the Wronskian of yb1HtL and yb2
HtL:WHtL �
ã3 tcosH4 tL ã3 t
sinH4 tLâ
âtIã3 t
cosH4 tLM â
âtIã3 t
sinH4 tLM �
ã3 tcosH4 tL ã3 t
sinH4 tL3 ã3 t
cosH4 tL - 4 ã3 tsinH4 tL 4 ã3 t
cosH4 tL + 3 ã3 tsinH4 tL � 4 ã
6 t
Let f HtL � 2 sinI t
2M - cosI t
2M:
Let v1HtL � -à f HtL yb2HtL
WHtL â t and v2HtL � à f HtL yb1HtL
WHtL â t:
The particular solution will be given by:
ypHtL � v1HtL yb1HtL + v2HtL yb2
HtL
Compute v1HtL:v1HtL � -à
I-cosI t
2M + 2 sinI t
2MM sinH4 tL
4 ã3 tâ t
�39 cosI 7 t
2M - 119 cosI 9 t
2M - 156 sinI 7 t
2M + 68 sinI 9 t
2M
2652 ã3 t
Compute v2HtL:v2HtL � à
cosH4 tL I-cosI t
2M + 2 sinI t
2MM
4 ã3 tâ t �
-
-156 cosI 7 t
2M + 68 cosI 9 t
2M - 39 sinI 7 t
2M + 119 sinI 9 t
2M
2652 ã3 t
The particular solution is thus:
ypHtL � v1HtL yb1HtL + v2HtL yb2
HtL �
cosH4 tL I39 cosI 7 t
2M - 119 cosI 9 t
2M - 156 sinI 7 t
2M + 68 sinI 9 t
2MM
2652
-
I-156 cosI 7 t
2M + 68 cosI 9 t
2M - 39 sinI 7 t
2M + 119 sinI 9 t
2MM sinH4 tL
2652
Simplify:
ypHtL � -4
663
5 cos
t
2
- 14 sin
t
2
The general solution is given by:
Answer:
yHtL � ycHtL + ypHtL �
c1 ã3 t
cosH4 tL + c2 ã3 t
sinH4 tL -4
663
5 cos
t
2
- 14 sin
t
2
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