Resuelva por coeficientes indeterminados y variacion de parametros by Gerson Villa Gonzalez

ex Resuelva y '' 2y ' y  x Solución Aquí n  2 y yh  c1e  c 2xe por lo tanto x

yp  v1e x  v 2 xe x x x Dado que y1  e ,y2  xe y (x)  e / x de la ecuación tenemos x

v '1 y1  v '2 y 2  0

  v '1 y '1  v '2 y '2  (x) 

 

v '1 e x  v '2  xe   0

   v ' e

v '1 e

 xe



ex  x

Resolviendo este conjunto de ecuaciones, obtenemos v '1  1 y v'2  1/ x . De este modo,

v1   v '1 dx   1dx   x 1 v 2   v '2 dx   dx  ln x x Sustituyendo estos valores obtenemos

yp   xe x  xe x ln x Por lo tanto, la solución general es

y  yc  yp  c1e x  c 2 xe x  xe x  xe x ln x  y  c1e x  c 3 xe x  xe x ln x  c 3  c 2  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 + ¢¢

y ¢¢ HxL 2 y ¢ HxL - yHxL +

ãx x ãx x

Alternate form assuming x is positive:

ãx x Iy ¢¢ HxL - 2 y ¢ HxL + yHxLM

Differential equation solution: x

Approximate form

Step-by-step solution

yHxL c1 ã + c2 ã x + ã x logHxL logHxL is the natural logarithm »

Plots of sample individual solutions:

yH1L 1

y¢

yH1L 0

y¢

y ¢ H1L 0

y ¢ H1L 1

Generated by Wolfram|Alpha (www.wolframalpha.com) on March 23, 2014 from Champaign, IL. © Wolfram Alpha LLC— A Wolfram Research Company

y'' -2y'+y=e^x/x

Sample solution family: y 60 50

Hsampling yH1L and y ¢ H1LL

40 30 20 10 x 1.5

2.0

2.5

3.0

Possible Lagrangian:

LIy ¢ , y, xM

-ã-2 x y 2 + ã-2 x y ¢ 2 +

2 ã-x y x

Generated by Wolfram|Alpha (www.wolframalpha.com) on March 23, 2014 from Champaign, IL. © Wolfram Alpha LLC— A Wolfram Research Company

Solve -2

â yHxL

âx

â2 yHxL âx

ãx x

+ yHxL

The general solution will be the sum of the complementary solution and particular solution. Find the complementary solution by solving

â2 yHxL âx2

-2

â yHxL âx

+ yHxL 0:

Assume a solution will be proportional to ãΛ x for some constant Λ. Substitute yHxL ãΛ x into the differential equation: â2 âx

IãΛ x M - 2

Substitute

â âx

â2

IãΛ x M Λ2 ãΛ x and

âx2 Λx

Λ2 ãΛ x - 2 Λ ã

IãΛ x M + ãΛ x 0

+ ãΛ x 0

â IãΛ x M âx

Λ ãΛ x :

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 y1 HxL c1 ã x , y2 HxL c2 ã x x as solutions, where c1 and c2 are arbitrary constants.

The general solution is the sum of the above solutions: yHxL y1 HxL + y2 HxL c1 ã x + c2 ã x x

Determine the particular solution to

â2 yHxL âx2

+ yHxL - 2

â yHxL âx

parameters:

List the basis solutions in yc HxL: yb1 HxL ã x and yb2 HxL ã x x

Compute the Wronskian of yb1 HxL and yb2 HxL: ãx

WHxL

â Hã x L âx

Let f HxL

ãx x

â Hã x âx

ãx ãx x ã2 x x x ã ã + ãx x

ãx : x

Let v1 HxL -à

f HxL yb2 HxL WHxL

â x and v2 HxL à

f HxL yb1 HxL WHxL

â x:

The particular solution will be given by: y p HxL v1 HxL yb1 HxL + v2 HxL yb2 HxL

Compute v1 HxL:

v1 HxL -à 1 â x -x

Compute v2 HxL: 1 v2 HxL à â x logHxL x The particular solution is thus:

y p HxL v1 HxL yb1 HxL + v2 HxL yb2 HxL -Hã x xL + ã x x logHxL

Simplify:

y p HxL ã x x HlogHxL - 1L

The general solution is given by:

yHxL yc HxL + y p HxL c1 ã x + c2 ã x x + ã x x HlogHxL - 1L

Simplify the arbitrary constants: Answer:

yHxL ã x x logHxL + c1 ã x + c2 ã x x

ãx x

by variation of

t t   cos   2 2

Resuelva y '' 6y ' 25y  2sen  Solución La ecuación característica es

2  6  25  0 Usando la formula cuadrática encontramos que sus raíces son:



  6  

 6 

 4  25 

 3  i4

Estas raíces son un par conjugado complejo, de modo que la solución general es:

y  c1e3t cos4t  c 2e3tsen4t Aquí (t) tiene la forma siguiente, con la variable independiente t reemplazando a x,k1  2,k 2  1 y  

1 (x)  k1senx  k 2 cos x donde k1,k 2 y  son 2

constantes conocidas. Se asume una solución de la forma

yp  Asenx  Bcos x Donde A y B son constantes a ser determinadas Donde tenemos

t t yp  Asen    Bcos   2 2 A t B t y 'p  cos    sen   2 2 2 2 y ''p  

A t B t sen    cos   4 2 4 2

Sustituyendo los resultados en la ecuación diferencial obtenemos

 A A  t B  t  t B  t  t  t    4 sen  2   4 cos  2    6  2 cos  2   2 sen  2    25  Asen  2   Bcos  2                  t t  2sen    cos   2 2 O de manera equivalente

99   99  t  t t t A  3B  sen     3A  B  cos    2sen    cos    4   4  2   2  2  2 Igualando los coeficientes de los términos similares tenemos

99 99 A  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

yp 

56  t  20 t sen    cos   663  2  663  2

Y la solución general es

y  yc  yp  c1e3t cos 4t  c 2e3t sen4t 

56  t  20 t sen    cos   663  2  663 2

y'' -6y'+25y=2senHt 2L-cosHt 2L

Input:

y ¢¢ HtL - 6 y ¢ HtL + 25 yHtL 2 sin

t - cos

ODE classification:

second-order linear ordinary differential equation Alternate forms:

y ¢¢ HtL + 25 yHtL + cos

t 2

6 y ¢ HtL + 2 sin

y ¢¢ HtL 6 y ¢ HtL - 25 yHtL + 2 sin y ¢¢ HtL - 6 y ¢ HtL + 25 yHtL -

t 2 t

- cos

+ä ã

2 1

ät 2

ät

+ä ã2 2

Differential equation solution:

Approximate form

yHtL c1 ã3 t sinH4 tL + c2 ã3 t cosH4 tL +

663

20 -

sin 2

Step-by-step solution

t cos

663

Plots of sample individual solutions:

yH0L 1

y¢

yH0L 0

y¢

y ¢ H0L 0

y ¢ H0L 1

Generated by Wolfram|Alpha (www.wolframalpha.com) on March 23, 2014 from Champaign, IL. © Wolfram Alpha LLC— A Wolfram Research Company

y'' -6y'+25y=2sen(t/2)-cos(t/2)

Sample solution family: y 100 80

Hsampling yH0L and y ¢ H0LL

60 40 20 t 0.2

0.4

0.6

0.8

1.0

1.2

-20

Possible Lagrangian:

LIy ¢ , y, tM

-25 ã-6 t y 2 + ã-6 t y ¢ 2 + 2 ã-6 t y 2 sin

t - cos

Generated by Wolfram|Alpha (www.wolframalpha.com) on March 23, 2014 from Champaign, IL. © Wolfram Alpha LLC— A Wolfram Research Company

Solve -6

â yHtL

ât

â2 yHtL ât2

+ 25 yHtL 2 sinI 2 M - cosI 2 M : t

The general solution will be the sum of the complementary solution and particular solution. Find the complementary solution by solving

â2 yHtL ât2

-6

â yHtL ât

+ 25 yHtL 0:

Assume a solution will be proportional to ãΛ t for some constant Λ. Substitute yHtL ãΛ t into the differential equation: â2 ât

IãΛ t M - 6

Substitute

â ât

â2

IãΛ t M + 25 ãΛ t 0

IãΛ t M Λ2 ãΛ t and

â IãΛ t M ât

ât2 Λt

Λ2 ãΛ t - 6 Λ ã + 25 ãΛ t 0

Λ ãΛ t :

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 y1 HtL c1 ãH3+4 äL t , y2 HtL 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 y1 HtL + y2 HtL c1 ãH3+4 äL t + c2 ãH3-4 äL t

Apply Euler's identity ãΑ+ä Β ãΑ cosHΒL + ä ãΑ sinHΒL:

yHtL c1 Iã3 t cosH4 tL + ä ã3 t sinH4 tLM + c2 Iã3 t cosH4 tL - ä ã3 t sinH4 tLM

Regroup terms:

yHtL Hc1 + c2 L ã3 t cosH4 tL + ä Hc1 - c2 L ã3 t sinH4 tL

Redefine c1 + c2 as c1 and ä Hc1 - c2 L as c2 , since these are arbitrary constants: yHtL c1 ã3 t cosH4 tL + c2 ã3 t sinH4 tL

Determine the particular solution to

â2 yHtL

+ 25 yHtL - 6

ât2

â yHtL ât

variation of parameters:

List the basis solutions in yc HtL:

2 sinI 2 M - cosI 2 M by t

yb1 HtL ã3 t cosH4 tL and yb2 HtL ã3 t sinH4 tL

Compute the Wronskian of yb1 HtL and yb2 HtL: ã3 t cosH4 tL

â Iã3 t ât

WHtL

ã3 t sinH4 tL

â Iã3 t ât

cosH4 tLM

ã3 t cosH4 tL

sinH4 tLM

ã3 t sinH4 tL

3 ã3 t cosH4 tL - 4 ã3 t sinH4 tL 4 ã3 t cosH4 tL + 3 ã3 t sinH4 tL Let f HtL 2 sinI 2 M - cosI 2 M: t

f HtL yb2 HtL

Let v1 HtL -à

4 ã6 t

WHtL

f HtL yb1 HtL

â t and v2 HtL à

WHtL

â t:

The particular solution will be given by: y p HtL v1 HtL yb1 HtL + v2 HtL yb2 HtL

Compute v1 HtL: v1 HtL -à

I-cosI 2 M + 2 sinI 2 MM sinH4 tL

39 cosI

4 ã3 t

ât

7t 9t 7t 9t M - 119 cosI 2 M - 156 sinI 2 M + 68 sinI 2 M 2

2652 ã3 t

Compute v2 HtL: v2 HtL à

cosH4 tL I-cosI 2 M + 2 sinI 2 MM

-156 cosI

4 ã3 t

ât

7t 9t 7t 9t M + 68 cosI 2 M - 39 sinI 2 M + 119 sinI 2 M 2

2652 ã3 t

The particular solution is thus:

y p HtL v1 HtL yb1 HtL + v2 HtL yb2 HtL cosH4 tL I39 cosI I-156 cosI

7t 9t 7t 9t M - 119 cosI 2 M - 156 sinI 2 M + 68 sinI 2 MM 2

2652

7t 9t 7t 9t M + 68 cosI 2 M - 39 sinI 2 M + 119 sinI 2 MM sinH4 tL 2

2652

Simplify: y p HtL -

4 5 cos 663

t - 14 sin

The general solution is given by: Answer:

yHtL yc HtL + y p HtL

c1 ã3 t cosH4 tL + c2 ã3 t sinH4 tL -

4 5 cos 663

t - 14 sin