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