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UNIVERSIDAD NACIONAL DE JAÉN CARRERA PROFESIONAL EN INGENIERIA DE INDUSTRIAS ALIMENTARIAS

TEMA: ECUACIONES DIFERENCIALES DE ORDEN SUPERIOR

DOCENTE: LIC.MAT. ROMAN MEDINA AGUILAR.

CATEDRA: ECUACIONES

INTEGRANTES: BRAVO LLACSAHUANGA MAYCOL GUEVARA REQUEJO CLAUDIA MELENDRES PEÑA JAIME SANCHEZ MARTINEZ TATIANA ZAMORA DIAZ IRIS

Jaén, Noviembre de 2017

1. Formar las ecuaciones diferenciales lineales homogéneas si se conocen las raíces de sus ecuaciones características y escribir sus soluciones generales. a. r1 =1; r2 =2; r3 =3

r - 1= 0

v

r - 2= 0

v

r - 3= 0

(𝑟 − 1) (𝑟 − 2) (𝑟 − 3) = 0 𝑟 3 − 6𝑟 2 + 11𝑟 − 6 = 0 𝑬. 𝑫: 𝑦 𝑖𝑖𝑖 − 6𝑦 𝑖𝑖 + 11𝑦 𝑖 − 6𝑦 = 0 𝑺𝒐𝒍. 𝑩á𝒔𝒊𝒄𝒂: 𝑒 𝑥 ; 𝑒 2𝑥 ; 𝑒 3𝑥 𝒀𝒈 = 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 2𝑥 + 𝐶3 𝑒 3𝑥

b. r1 =1; r2 =2; r3 =2 r–1=0

v

r–2=0

v

r–2=0

(𝑟 − 1) (𝑟 − 2) (𝑟 − 2) = 0 𝑟 3 − 5𝑟 2 + 8𝑟 − 4 = 0 𝑬. 𝑫: 𝑦 𝑖𝑖𝑖 − 5𝑦 𝑖𝑖 + 8𝑦 𝑖 − 4𝑦 = 0 𝑺𝒐𝒍. 𝑩á𝒔𝒊𝒄𝒂: 𝑒 𝑥 ; 𝑒 2𝑥 ; 𝑒 2𝑥 𝒀𝒈 = 𝐶1 𝑒 𝑥 + (𝐶2 + 𝐶3 )𝑒 2𝑥

c. r1 =1; r2 =1; r3 =1 r–1=0

v

r–1=0

v

(𝑟 − 1)3 = 0 𝑟 3 − 3𝑟 2 + 3𝑟 − 1 = 0 𝑬. 𝑫: 𝑦 𝑖𝑖𝑖 − 3𝑦 𝑖𝑖 + 3𝑦 𝑖 − 𝑦 = 0 𝑺𝒐𝒍. 𝑩á𝒔𝒊𝒄𝒂: 𝑒 𝑥 ; 𝑒 𝑥 ; 𝑒 𝑥 𝒀𝒈 = (𝐶1 + 𝐶2 + 𝐶3 )𝑒 𝑥

r–1=0

d. r1 =0; r2 =2+3i; r3 =2-3i r–1=0

v

r + 2 + 3i = 0

v

r + 2 – 3i = 0

𝑟(𝑟 + 2 + 3𝑖)(𝑟 + 2 − 3𝑖) = 0 𝑟[(𝑟 + 2)2 − (3𝑖)2 ] = 0 𝑟(𝑟 2 + 4𝑟 + 4 − 3) = 0 𝑟 3 + 4𝑟 2 + 𝑟 = 0 𝑬. 𝑫: 𝑦 𝑖𝑖𝑖 + 4𝑦 𝑖𝑖 + 𝑦 𝑖 = 0 𝑺𝒐𝒍. 𝑩á𝒔𝒊𝒄𝒂: 𝑒 0𝑥 ; 𝑒 2𝑥 𝑐𝑜𝑠3𝑥; 𝑒 2𝑥 𝑠𝑒𝑛3𝑥 𝒀𝒈 = 𝐶1 + 𝐶2 𝑒 2𝑥 𝑐𝑜𝑠3𝑥 + 𝐶3 𝑒 2𝑥 𝑠𝑒𝑛3𝑥

2. Formar las ecuaciones diferenciales lineales homogéneas si se dan sus soluciones básicas de solución. a. e-2x, e3x 𝑟 = −2

𝒗

𝑟=3

𝑟+2=0

𝒗

𝑟−3 =0

(𝑟 + 2) (𝑟 − 3) = 0 𝑟 2 − 3𝑟 + 2𝑟 − 6 = 0 𝑟2 − 𝑟 − 6 = 0 𝑬. 𝑫: 𝑦 𝑖𝑖 − 𝑦 𝑖 − 6𝑦 = 0 b. 1, x e0x; xe0x 𝑟 = −𝑖

𝒗

𝑟+𝑖 = 0

𝑟=𝑖 𝒗

; 𝒊𝒆𝑪

𝑟−𝑖 = 0

(𝑟 + 𝑖) (𝑟 − 𝑖) = 0 𝑟2 − 𝑖2 = 0 𝑟2 + 1 = 0 𝑬. 𝑫: 𝑦 𝑖𝑖 − 𝑦 = 0

c. ex, xex, x2ex 𝑟=1

𝒗

𝑟−1=0

𝒗

𝑟=1

𝒗

𝑟−1=0

𝑟=1 𝒗

𝑟−1=0

(𝑟 − 1) (𝑟 − 1)(𝑟 − 1) = 0 (𝑟 2 − 𝑟 − 𝑟 + 1)(𝑟 − 1) = 0 (𝑟 2 − 2𝑟 + 1)(𝑟 − 1) = 0 𝑟 3 − 𝑟 2 − 2𝑟 2 + 2𝑟 + 𝑟 − 1 = 0 𝑟 3 − 3𝑟 2 + 3𝑟 − 1 = 0 𝑬. 𝑫: 𝑦 𝑖𝑖𝑖 − 3𝑦 𝑖𝑖 + 3𝑦 𝑖 − 𝑦 = 0 d. e2x, excos3x, exsen3x 𝑟=2

𝒗

𝑟−2=0

𝑟 = 1 + 3𝑖 𝒗

𝒗

𝑟 = 1 − 3𝑖 ; 𝒊𝒆𝑪

𝑟 − 1 − 3𝑖 = 0

𝒗

𝑟 − 1 + 3𝑖 = 0

(𝑟 − 2) (𝑟 − 1 − 3𝑖)(𝑟 − 1 + 3𝑖) = 0 (𝑟 − 𝑟 − 3𝑖𝑟 − 2𝑟 + 2 + 6𝑖)(𝑟 − 1 + 3𝑖) = 0 (𝑟 2 − 3𝑟 − 3𝑟𝑖 + 2 + 6𝑖)(𝑟 − 1 + 3𝑖) = 0 𝑟 3 − 3𝑟 2 − 3𝑖𝑟 2 + 2𝑟 + 6𝑖𝑟 − 𝑟 2 + 3𝑟 + 3𝑖𝑟 − 2 − 6𝑖 + 3𝑖𝑟 2 − 9𝑖𝑟 − 9𝑖 2 𝑟 + 6𝑖 + 18𝑖 2 = 0 𝑟 3 − 4𝑟 2 + 5𝑟 − 9𝑟𝑖 2 + 18𝑖 2 = 0 𝑟 3 − 4𝑟 2 + 5𝑟 − 9𝑟 + 18 = 0 𝑟 3 − 4𝑟 2 − 4𝑟 + 18 = 0 2

𝑬. 𝑫: 𝑦 𝑖𝑖𝑖 − 4𝑦 𝑖𝑖 − 4𝑦 𝑖 + 18𝑦 = 0 e. 1, e-2xcos3x, e-2xsen3x 𝑟=0 𝑟=0

𝒗 𝒗

𝑟 = −2 + 3𝑖

𝒗

𝑟 + 2 − 3𝑖 = 0

𝑟 = −2 − 3𝑖 ; 𝒊𝒆𝑪 𝒗

𝑟 + 2 + 3𝑖 = 0

(𝑟 − 2)(𝑟 + 2 − 3𝑖)(𝑟 + 2 − 3𝑖) = 0 𝑟(𝑟 + 2𝑟 + 3𝑟𝑖 + 2𝑟 + 4 + 6𝑖 − 3𝑟𝑖 − 6𝑖 − 9𝑖 2 ) = 0 𝑟(𝑟 2 + 4𝑟 + 4 − 9𝑖 2 ) = 0 𝑟 3 + 4𝑟 2 + 4𝑟 − 9𝑟(−1) = 0 𝑟 3 + 4𝑟 2 + 4𝑟 + 9𝑟 = 0 𝑟 3 + 4𝑟 2 + 13𝑟 = 0 2

𝑬𝑫: 𝑦 𝑖𝑖𝑖 + 4𝑦 𝑖𝑖 + 13𝑦′ = 0

f. Sen5x, cos5x 𝑟 = −5𝑖

𝒗

𝑟 = +5𝑖 ; 𝒊𝒆𝑪

(𝑟 + 5𝑖) = 0; (𝑟 − 5𝑖) = 0 (𝑟 + 5𝑖)(𝑟 − 5𝑖) = 0 2 (𝑟 − 5𝑟𝑖 + 5𝑟𝑖 − 25𝑖 2 ) = 0 (𝑟 2 − 25(−1)) = 0 𝑟 2 + 25 = 0

𝑬𝑫: 𝑦 ′′ + 25𝑦 = 0

3. Resolver las siguientes ecuaciones diferenciales lineales homogéneas de coeficientes constantes: a. yiii - 2yii + 2yi = 0 𝑷. 𝑪: 𝑃(𝑟) = 𝑟 3 − 2𝑟 2 + 2𝑟 𝑃(𝑟) = 0 𝑟 3 − 2𝑟 2 + 2𝑟 = 0 𝑟(𝑟 2 − 2𝑟 + 2) = 0 𝑟[(𝑟 − 1)2 − 𝑖 2 ] = 0 𝑟(𝑟 − 1 − 𝑖)(𝑟 − 1 + 𝑖) = 0 𝑟=0

𝒗

𝑟 = 1+𝑖

𝒗

𝑟 =1−𝑖

𝑺𝒐𝒍. 𝑩á𝒔𝒊𝒄𝒂: 𝑒 0𝑥 ; 𝑒 𝑥 𝑐𝑜𝑠𝑥; 𝑒 𝑥 𝑠𝑒𝑛𝑥 𝒀𝒈 = 𝐶1 + 𝐶2 𝑒 𝑥 𝑐𝑜𝑠𝑥 + 𝐶3 𝑒 𝑥 𝑠𝑒𝑛𝑥

b. yiv + 2yiii + 4yii – 2yi – 5y = 0 𝑷. 𝑪: 𝑃(𝑟) = 𝑟 4 + 2𝑟 3 + 4𝑟 2 − 2𝑟 − 5 𝑃(𝑟) = 0 𝑟 4 + 2𝑟 3 + 4𝑟 2 − 2𝑟 − 5 = 0 (𝑟 − 1)(𝑟 + 1)(𝑟 2 + 2𝑟 + 5) = 0 (𝑹𝒖𝒇𝒇𝒊𝒏𝒊) (𝑟 − 1)(𝑟 + 1)[(𝑟 + 1)2 + (2𝑖)2 ] = 0 ; 𝒊𝒆𝑪 (𝑟 − 1)(𝑟 + 1)(𝑟 + 1 + 2𝑖)(𝑟 + 1 − 2𝑖) = 0 𝑟=1

𝒗

𝑟 = −1

𝒗

𝑟 = −1 − 2𝑖

𝒗

𝑟 = −1 + 2𝑖

𝑺𝒐𝒍. 𝑩á𝒔𝒊𝒄𝒂: 𝑒 𝑥 ; 𝑒 −𝑥 , 𝑒 −𝑥 𝑐𝑜𝑠2𝑥; 𝑒 −𝑥 𝑠𝑒𝑛2𝑥 𝒀𝒈 = 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 −𝑥 + 𝐶3 𝑒 −𝑥 𝑐𝑜𝑠2𝑥 + 𝐶4 𝑒 −𝑥 𝑠𝑒𝑛2𝑥

Ruffini:

1 1 1 -1 1

2 1 3 -1 2

4 3 7 -2 5

-2 7 5 -5 0

-5 5 0

(𝑟 − 1)(𝑟 + 1)(𝑟 2 + 2𝑟 + 5)

c. yv + 2yiii + 10yii + yi + 10y = 0 𝑷. 𝑪: 𝑃(𝑟) = 𝑟 5 + 2𝑟 3 + 10𝑟 2 + 𝑟 + 10 𝑃(𝑟) = 0 𝑟 5 + 2𝑟 3 + 10𝑟 2 + 𝑟 + 10 = 0 (𝑟 + 2)(𝑟 4 + 2𝑟 3 + 6𝑟 2 − 2𝑟 + 5) = 0 (𝑹𝒖𝒇𝒇𝒊𝒏𝒊) (𝑟 + 2)(𝑟 2 + 1)(𝑟 2 − 2𝑟 + 5) = 0 (𝒂𝒔𝒑𝒂 𝒅𝒐𝒃𝒍𝒆 𝒆𝒔𝒑𝒆𝒄𝒊𝒂𝒍) (𝑟 − 1)(𝑟 + 1)(𝑟 + 1 + 2𝑖)(𝑟 + 1 − 2𝑖) = 0 𝑟 = −2

𝒗

𝑟 = 0+𝑖

𝒗

𝑟 = 0−𝑖

𝒗

𝑟 = 1 + 2𝑖

𝒗

𝑟 = 1 − 2𝑖

𝑺𝒐𝒍. 𝑩á𝒔𝒊𝒄𝒂: 𝑒 −2𝑥 ; 𝑒 0𝑥 𝑐𝑜𝑠𝑥, 𝑒 0𝑥 𝑠𝑒𝑛𝑥, 𝑒 𝑥 𝑐𝑜𝑠2𝑥; 𝑒 𝑥 𝑠𝑒𝑛2𝑥 𝒀𝒈 = 𝐶1 𝑒 −2𝑥 + 𝐶2 𝑐𝑜𝑠𝑥 + 𝐶3 𝑠𝑒𝑛𝑥 + 𝐶4 𝑒 𝑥 𝑐𝑜𝑠𝑥 + 𝐶5 𝑒 𝑥 𝑠𝑒𝑛𝑥 𝒀𝒈 = 𝐶1 𝑒 −2𝑥 + (𝐶2 + 𝐶4 𝑒 𝑥 )𝑐𝑜𝑠𝑥 + (𝐶3 + 𝐶5 𝑒 𝑥 )𝑠𝑒𝑛𝑥

Ruffini:

1

0 -2 -2

-2 1

2 4 6

10 -12 -2

1 4 5

10 -10 0

d. yviii + 8yiv + 16y = 0 𝑷. 𝑪: 𝑃(𝑟) = 𝑟 8 + 8𝑟 4 + 16 𝑃(𝑟) = 0 𝑟 8 + 8𝑟 4 + 16 = 0 (𝑟 4 + 4)2 = 0 (𝑟 2 + 2𝑖)2 (𝑟 2 − 2𝑖)2 = 0 2 [𝑟 − (1 − 𝑖)2 ]2 [𝑟 2 − (1 + 𝑖)2 ]2 = 0 [(𝑟 + (1 − 𝑖))2 (𝑟 − (1 − 𝑖))2 ] [(𝑟 − (1 + 𝑖))2 (𝑟 + (1 + 𝑖))2 ] = 0 𝑟1,2 = −1 + 𝑖

𝒗

𝑟3,4 = 1 − 𝑖

𝒗

𝑟5,6 = 1 + 𝑖

𝒗

𝑟7,8 = −1 − 𝑖

𝑺𝒐𝒍. 𝑩á𝒔𝒊𝒄𝒂: 𝑒 −𝑥 𝑐𝑜𝑠𝑥; 𝑒 −𝑥 𝑠𝑒𝑛𝑥; 𝑥𝑒 −𝑥 𝑐𝑜𝑠𝑥; 𝑥𝑒 −𝑥 𝑠𝑒𝑛𝑥; 𝑒 𝑥 𝑐𝑜𝑠𝑥; 𝑒 𝑥 𝑠𝑒𝑛𝑥; 𝑥𝑒 𝑥 𝑐𝑜𝑠𝑥; 𝑥𝑒 𝑥 𝑠𝑒𝑛𝑥 𝒀𝒈 = 𝐶1 𝑒 −𝑥 𝑐𝑜𝑠𝑥 + 𝐶2 𝑒 −𝑥 𝑠𝑒𝑛𝑥 + 𝐶3 𝑥𝑒 −𝑥 𝑐𝑜𝑠𝑥 + 𝐶4 𝑥𝑒 −𝑥 𝑠𝑒𝑛𝑥 + 𝐶5 𝑒 𝑥 𝑐𝑜𝑠𝑥 𝐶6 𝑒 𝑥 𝑠𝑒𝑛𝑥 + 𝐶7 𝑥𝑒 𝑥 𝑐𝑜𝑠𝑥 + 𝐶8 𝑥𝑒 𝑥 𝑠𝑒𝑛𝑥 𝒀𝒈 = [(𝐶1 + 𝐶3 𝑥)𝑐𝑜𝑠𝑥 + (𝐶2 + 𝐶4 𝑥)𝑠𝑒𝑛𝑥]𝑒 −𝑥 + [(𝐶5 + 𝐶7 𝑥)𝑐𝑜𝑠𝑥 + (𝐶6 + 𝐶8 𝑥)𝑠𝑒𝑛𝑥]𝑒 𝑥 e. yvi + y = 0 𝑷. 𝑪; 𝑷(𝒓) = 𝒓𝟔 + 𝟏 = 𝟎 𝒓𝟔 + 𝟏 = 𝟎 ; 𝒊𝟐 = −𝟏 (𝒓𝟐 + 𝟏)𝟑 = 𝟎 (𝒓𝟑 )𝟐 − 𝒊𝟐 = 𝟎 ; 𝒊€ 𝒄 (𝒓𝟑

𝑖 2 = −1 𝟑 − 𝒊)(𝒓 𝒊) = 𝟎 ; 𝑖 3 =+−𝑖

(𝒓𝟑 + 𝒊𝟑 )(𝒓𝟑 − 𝒊𝟑 ) = 𝟎

(𝒓 + 𝒊)(𝒓𝟐 − 𝒊𝒓 + 𝒊𝟐 )(𝒓 − 𝒊)(𝒓𝟐 + 𝒊𝒓 + 𝒊𝟐 ) = 𝟎 (𝒓 + 𝒊)(𝒓𝟐 − 𝒊𝒓 − 𝟏)(𝒓 − 𝒊)(𝒓𝟐 + 𝒊𝒓 − 𝟏) = 𝟎 𝒓𝟏 = −𝒊; 𝒓𝟐,𝟑

𝒓𝟏 = −𝒊

𝒗

𝒊 ± √𝒊𝟐 + 𝟒 √𝟑 𝒊 = ; 𝒓𝟒 = 𝒊; 𝒓𝟒 = − ; 𝒓𝟓,𝟔 𝟐 𝟐 𝟐 𝟐 −𝒊 ± √𝒊 + 𝟒 = 𝟐 𝒓𝟐 =

=𝒊

√𝟑 𝒊 + 𝟐 𝟐 𝒗

𝒓𝟓

√𝟑 𝒊 + 𝒗 𝟐 𝟐 √𝟑 𝒊 √𝟑 𝒊 = − ; 𝒓𝟔 = − − 𝟐 𝟐 𝟐 𝟐 𝒗

𝒓𝟑 = −

𝒓𝟒

−√𝟑 √𝟑 √𝟑 √𝟑 𝒙 𝒙 𝒙 𝒙 𝑺𝒐𝒍. 𝑩á𝒔𝒊𝒄𝒂: 𝒔𝒆𝒏𝒙, 𝒆 𝟐 𝒄𝒐𝒔 ( ) , 𝒆 𝟐 𝒄𝒐𝒔 ( ) , 𝒄𝒐𝒔𝒙, 𝒆 𝟐 𝒔𝒆𝒏( ), 𝒆− 𝟐 𝒔𝒆𝒏( ) 𝟐 𝟐 𝟐 𝟐 −√𝟑 √𝟑 √𝟑 𝒙 𝒙 𝒙 𝒚(𝒈) = 𝑪𝟏 𝒔𝒆𝒏𝒙 + 𝑪𝟐 𝒆 𝟐 𝒄𝒐𝒔 ( ) + 𝑪𝟑 𝒆 𝟐 𝒄𝒐𝒔 ( ) + 𝑪𝟒 𝒄𝒐𝒔𝒙 + 𝑪𝟓 𝒆 𝟐 𝒔𝒆𝒏 ( ) 𝟐 𝟐 𝟐 √𝟑 𝒙 + 𝑪𝟔 𝒆− 𝟐 𝒔𝒆𝒏( ) 𝟐 𝒙 𝒙 √𝟑 𝒙 𝒚(𝒈) = 𝑪𝟏 𝒔𝒆𝒏𝒙 + 𝑪𝟒 𝒄𝒐𝒔 𝒙 + (𝑪𝟐 𝒄𝒐𝒔 ( ) + 𝑪𝟓 𝒔𝒆𝒏( ))𝒆 𝟐 𝒙 + (𝑪𝟑 𝒄𝒐𝒔 ( ) 𝟐 𝟐 𝟐 𝒙 −√𝟑𝒙 + 𝑪𝟔 𝒔𝒆𝒏( ))𝒆 𝟐 𝟐

f. yiii + 5yii + 17yi + 13y = 0; y(0) = 0; yi(0) = 1; yii(0) = 6 𝑷. 𝑪: 𝑃(𝑟) = 𝑟 3 + 5𝑟 2 + 17𝑟 + 13 𝑃(𝑟) = 0 𝑟 3 + 5𝑟 2 + 17𝑟 + 13 = 0 (𝑟 + 1)(𝑟 2 + 4𝑟 + 13) = 0 (𝑹𝒖𝒇𝒇𝒊𝒏𝒊) (𝑟 + 1)(𝑟 + 2)2 + (3𝑖)2 = 0 𝑟 = −1

𝒗

𝑟 = −2 + 3𝑖

𝒗

𝑟 = −2 + 3𝑖

𝑺𝒐𝒍. 𝑩á𝒔𝒊𝒄𝒂: 𝑒 −𝑥 ; 𝑒 −2𝑥 𝑐𝑜𝑠3𝑥, 𝑒 −2𝑥 𝑠𝑒𝑛3𝑥 𝒀𝒈 = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 −2𝑥 𝑐𝑜𝑠3𝑥 + 𝐶3 𝑒 −2𝑥 𝑠𝑒𝑛3𝑥

i.

Encontrar el C1: 𝒚(𝟎) = 𝟎 𝒀(𝒙)𝒈 = 𝐶1 𝑒 + 𝐶2 𝑒 −2𝑥 𝑐𝑜𝑠3𝑥 + 𝐶3 𝑒 −2𝑥 𝑠𝑒𝑛3𝑥 𝒀(𝟎)𝒈 = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 −2𝑥 𝑐𝑜𝑠3𝑥 + 𝐶3 𝑒 −2𝑥 𝑠𝑒𝑛3𝑥 𝒀(𝟎)𝒈 = 𝐶1 (1) + 𝐶2 (1)(1) 𝟎 = 𝐶1 + 𝐶2 𝑪𝟏 = −𝐶2 −𝑥

ii.

Encontrar el C2: 𝒚′(𝟎) = 𝟏 𝒀(𝒙)𝒈 = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 −2𝑥 𝑐𝑜𝑠3𝑥 + 𝐶3 𝑒 −2𝑥 𝑠𝑒𝑛3𝑥 𝒀′(𝒙)𝒈 = −𝐶1 𝑒 −𝑥 + 𝐶2 (−2𝑒 −2𝑥 𝑐𝑜𝑠3𝑥 − 𝑒 −2𝑥 𝑠𝑒𝑛3𝑥) + 𝐶3 (−2𝑒 −2𝑥 𝑠𝑒𝑛3𝑥 + 𝑒 −2𝑥 𝑐𝑜𝑠3𝑥) 𝒀′(𝟎)𝒈 = −𝐶1 + 𝐶2 (−2) + 𝐶3 𝑹𝒆𝒆𝒎𝒑𝒍𝒂𝒛𝒂𝒎𝒐𝒔 𝒆𝒏 𝑪𝟏: 𝟏 = 𝐶2 − 2𝐶2 + 𝐶3 𝟏 = −𝐶2 + 𝐶3 𝑪𝟐 = −1 + 𝐶3

iii.

Encontrar el C3: 𝒚′′(𝟎) = 𝟔 𝒀′(𝒙)𝒈 = −𝐶1 𝑒 −𝑥 + 𝐶2 (−2𝑒 −2𝑥 𝑐𝑜𝑠3𝑥 − 𝑒 −2𝑥 𝑠𝑒𝑛3𝑥) + 𝐶3 (−2𝑒 −2𝑥 𝑠𝑒𝑛3𝑥 + 𝑒 −2𝑥 𝑐𝑜𝑠3𝑥) 𝒀′′ (𝒙)𝒈 = 𝐶1 𝑒 −𝑥 + 𝐶2 (4𝑒 −2𝑥 𝑐𝑜𝑠3𝑥 + 2𝑒 −2𝑥 𝑠𝑒𝑛3𝑥 + 2𝑒 −2𝑥 𝑠𝑒𝑛3𝑥 − 𝑒 −2𝑥 𝑐𝑜𝑠3𝑥) + 𝐶3 (4𝑒 −2𝑥 𝑠𝑒𝑛3𝑥 − 2𝑒 −2𝑥 𝑐𝑜𝑠3𝑥 − 2𝑒 −2𝑥 𝑐𝑜𝑠3𝑥 − 𝑒 −2𝑥 𝑠𝑒𝑛3𝑥) 𝒀′′(𝟎)𝒈 = 𝐶1 + 𝐶2 (4 − 1) + 𝐶3 (−2 − 2) 𝑹𝒆𝒆𝒎𝒑𝒍𝒂𝒛𝒂𝒎𝒐𝒔 𝒆𝒏 𝑪𝟏: 𝒀′′(𝟎)𝒈 = −𝐶2 + 3𝐶2 − 4𝐶3 𝒀′′(𝟎)𝒈 = −𝐶2 + 3𝐶2 − 4𝐶3 𝟔 = 2𝐶2 − 4𝐶3 𝑹𝒆𝒆𝒎𝒑𝒍𝒂𝒛𝒂𝒎𝒐𝒔 𝒆𝒏 𝑪𝟐: 𝟔 = 2𝐶3 − 2 − 4𝐶3 𝟖 = −2𝐶3 −𝟒 = 𝐶3

iv.

Reemplazando C3 en C1 y C2: ∗ 𝑪𝟑 = −𝟒 ∗ 𝑪𝟐 = −𝟏 + 𝑪𝟑 = −𝟏 − 𝟒 = −𝟓 ∗ 𝑪𝟏 = −𝑪𝟐 = 𝟓

∴ 𝒀 = 𝟓𝒆−𝟐𝒙 − 𝟓𝒆−𝟐𝒙 𝒄𝒐𝒔𝟑𝒙 − 𝟒𝒆−𝟐𝒙 𝒔𝒆𝒏𝟑𝒙

4. Resolver las siguientes ecuaciones diferenciales lineales no homogéneas de coeficientes constantes: a. yii – 4yi = xe4x 𝑟 2 − 4𝑟 = 0 𝑟(𝑟 − 4) = 0 𝑟=0 ˅ 𝑟=4

Y(c)=𝑦(𝑔) = 𝐶1 𝑒 0𝑥 + 𝐶2 𝑒 4𝑥 𝑦(𝑔) = 𝐶1 + 𝐶2 𝑒 4𝑥

Hallamos yp similar a 𝑓(𝑥) = 𝑥𝑒 4𝑥 Caso II, 2 𝑦𝑝 = 𝑥𝑒 4𝑥 (𝐴𝑥 + 𝐵) 𝑦𝑝 = 𝑒 4𝑥 (𝐴𝑥 2 + 𝐵𝑥)

Hallamos A ˄ B; hallar yp'' ˄ yp' Si 𝑦𝑝 = 𝑒 4𝑥 (𝐴𝑥 2 + 𝐵𝑥)  𝒚′𝒑 = 𝒆𝟒𝒙 (𝟐𝑨𝒙 + 𝑩) + 𝟒𝒆𝟒𝒙 (𝑨𝒙𝟐 + 𝑩𝒙) 𝑦′𝑝 = 𝑒 4𝑥 (2𝐴𝑥 + 𝐵) + 𝑒 4𝑥 (4𝐴𝑥 2 + 4𝐵𝑥)

 𝒚′′𝒑 = [𝟐𝑨𝒆𝟒𝒙 + 𝟒(𝟐𝑨𝒙 + 𝟒𝑩)𝒆𝟒𝒙 ] + [(𝟖𝑨𝒙 + 𝟒𝑩)𝒆𝟒𝒙 + 𝟒(𝟒𝑨𝒙𝟐 + 𝟒𝑩𝒙)𝒆𝟒𝒙 ] 𝑦′′𝑝 = [2𝐴𝑒 4𝑥 + (8𝐴𝑥 + 4𝐵)𝑒 4𝑥 ] + [(8𝐴𝑥 + 4𝐵)𝑒 4𝑥 + (16𝐴𝑥 2 + 16𝐵𝑥)𝑒 4𝑥 ]

Por lo tanto: 𝑦 ′′ 𝑃 − 4𝑦 ′ 𝑃 = 𝑥𝑒 4𝑥 [2𝐴𝑒 4𝑥 + (8𝐴𝑥 + 4𝐵)𝑒 4𝑥 ] + [(8𝐴𝑥 + 4𝐵)𝑒 4𝑥 + (16𝐴𝑥 2 + 16𝐵𝑥)𝑒 4𝑥 ] − 4[𝑒 4𝑥 (2𝐴𝑥 + 𝐵) + 𝑒 4𝑥 (4𝐴𝑥 2 + 4𝐵𝑥)] = 𝑥𝑒 4𝑥 [2𝐴 + (8𝐴𝑥 + 4𝐵)] + [(8𝐴𝑥 + 4𝐵) + (16𝐴𝑥 2 + 16𝐵𝑥)] − [(8𝐴𝑥 + 4𝐵) + (16𝐴𝑥 2 + 16𝐵𝑥)] = 𝑥

8𝐴𝑥 + 2𝐴 + 4𝐵 = 𝑥

Calculando:  8𝐴 = 1 𝐴=

1 8

 2𝐴 + 4𝐵 = 0 1 2 ( ) + 4𝐵 = 0 8 1 4𝐵 = − 4 1 𝐵=− 16

Remplazamos en: 𝑦𝑝 = 𝑒 4𝑥 (𝐴𝑥 2 + 𝐵𝑥) 1 1 𝑦𝑝 = 𝑒 4𝑥 ( 𝑥 2 − 𝑥) 8 16

Hallamos: 𝑦𝑔 = 𝑦𝑐 + 𝑦𝑝 1 1 𝑦𝑔 = 𝐶1 + 𝐶2 𝑒 4𝑥 + 𝑥 2 𝑒 4𝑥 − 𝑥𝑒 4𝑥 8 16

b. 𝐲 ′′ − 𝟖𝐲 ′ + 𝟏𝟔 = (𝟏 − 𝐱)𝐞𝟒𝐱 𝑦 ′′ − 8𝑦 ′ = 𝑒 4𝑥 − 𝑥𝑒 4𝑥 − 16

Hallamos yc de: 𝑦 ′′ − 8𝑦 ′ = 0  E.C: 𝑟 2 − 8𝑟 = 0 𝑟(𝑟 − 8) = 0 𝑟=0 ˅ 𝑟=8  yc: 𝑦(𝑔) = 𝐶1 𝑒 0𝑥 + 𝐶2 𝑒 8𝑥 𝑦(𝑔) = 𝐶1 + 𝐶2 𝑒 8𝑥

Hallamos yp similar a 𝑓(𝑥) = (1 − 𝑥)𝑒 4𝑥

Caso II,2  𝑦𝑝 = 𝐴𝑥 + 𝑥𝑒 4𝑥 (𝐵𝑥 + 𝐶) + 𝐷𝑒 4𝑥 𝑦𝑝 = 𝐴𝑥 + 𝐵𝑥 2 𝑒 4𝑥 + 𝐶𝑥𝑒 4𝑥 + 𝐷𝑒 4𝑥

Hallamos A ˄ B; hallar yp'' ˄ yp' Si 𝐴𝑥 + 𝐵𝑥 2 𝑒 4𝑥 + 𝐶𝑥𝑒 4𝑥 + 𝐷𝑒 4𝑥  𝑦′𝑝 = 𝐴 + 2𝐵𝑥𝑒 4𝑥 + 4𝐵𝑥 2 𝑒 4𝑥 + 4𝐶𝑥𝑒 4𝑥 + 𝐶𝑒 4𝑥 + 4𝐷𝑒 4𝑥  𝑦′′𝑝 = [(8𝐵𝑥 + 2𝐵) + (8𝐵𝑥 + 16𝑥 2 )]𝑒 4𝑥 + [(16𝐶𝑥 + 4𝑐) + 4𝑐]𝑒 4𝑥 + 16𝐷𝑒 4𝑥

Por lo tanto: 𝑦 ′′ 𝑃 − 8𝑦 ′ 𝑃 = 𝑒 4𝑥 − 𝑥𝑒 4𝑥 − 16

 [(8𝐵𝑥 + 2𝐵) + (8𝐵𝑥 + 16𝑥 2 )]𝑒 4𝑥 + [(16𝐶𝑥 + 4𝑐) + 4𝑐]𝑒 4𝑥 + 16𝐷𝑒 4𝑥 − 8𝐴 − 16𝐵𝑥𝑒 4𝑥 − 32𝐵𝑥 2 𝑒 4𝑥 − 32𝐶𝑥𝑒 4𝑥 − 8𝐶𝑒 4𝑥 − 32𝐷𝑒 4𝑥 = 𝑒 4𝑥 − 𝑥𝑒 4𝑥 − 16 −16𝐵𝑥 2 𝑒 4𝑥 + (2𝐵 − 16𝐷)𝑒 4𝑥 − 16𝐶𝑥𝑒 4𝑥 − 8𝐴 = 𝑒 4𝑥 − 𝑥𝑒 4𝑥 − 16

Calculando:  −8𝐴 = −16 𝐴=2  −16𝐶 = −1 1 𝐶= 16  −16𝐵 = 0 𝐵=0  2𝐵 − 16𝐷 = 1 16𝐷 = −1 1 𝐷=− 16

Remplazamos en: 𝑦𝑝 = 𝐴𝑥 + 𝐵𝑥 2 𝑒 4𝑥 + 𝐶𝑥𝑒 4𝑥 + 𝐷𝑒 4𝑥 1

1

 𝑦𝑝 = 2𝑥 + 0𝑥 2 𝑒 4𝑥 + 16 𝑥𝑒 4𝑥 − 16 𝑒 4𝑥 𝑦𝑝 = 2𝑥 +

1 1 𝑥𝑒 4𝑥 − 𝑒 4𝑥 16 16

Hallamos 𝑦𝑔 = 𝑦𝑐 + 𝑦𝑝 1

1

 𝑦𝑔 = 𝐶1 + 𝐶2 𝑒 8𝑥 + 2𝑥 + 16 𝑥𝑒 4𝑥 − 16 𝑒 4𝑥

c. 𝐲 ′′ + 𝟐𝐲 ′ − 𝟑𝐲 = 𝟏 + 𝐱𝐞𝐱

Hallamos yc de: 𝑦 ′′ + 2𝑦 ′ − 3𝑦 = 0  E.C: 𝑟 2 + 2𝑟 − 3 = 0 (𝑟 − 1)(𝑟 + 2) = 0 𝑟 = 1 ˅ 𝑟 = −2  yc: 𝑦(𝑔) = 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 −2𝑥

Hallamos yp similar a 𝑓(𝑥) = 1 + 𝑥𝑒 𝑥 Caso II, 2  𝑦𝑝 = 𝐴𝑥 + 𝑥𝑒 𝑥 (𝐵𝑥 + 𝐶) 𝑦𝑝 = 𝐴𝑥 + (𝐵𝑥 2 + 𝐶𝑥)𝑒 𝑥

Hallamos A ˄ B ˄ C; hallar yp'' ˄ yp' Si 𝑦𝑝 = 𝐴𝑥 + (𝐵𝑥 2 + 𝐶𝑥)𝑒 𝑥  𝑦 ′ 𝑝 = 𝐴 + (2𝐵𝑥 + 𝐶)𝑒 𝑥 + (𝐵𝑥 2 + 𝐶𝑥)𝑒 𝑥  𝑦′′𝑝 = [2𝐵𝑒 𝑥 + (2𝐵𝑥 + 𝐶)𝑒 𝑥 ] + [(2𝐵𝑥 + 𝐶)𝑒 𝑥 + (𝐵𝑥 2 + 𝐶𝑥)𝑒 𝑥 ] 𝑦′′𝑝 = (2𝐵 + 2𝐵𝑥 + 𝐶 + 2𝐵𝑥 + 𝐶 + 𝐵𝑥 2 + 𝐶𝑥)𝑒 𝑥 𝑦′′𝑝 = (𝐵𝑥 2 + 4𝐵𝑥 + 𝐶𝑥 + 2𝐵 + 2𝐶)𝑒 𝑥 

Por lo tanto: 𝑦 ′′ 𝑃 + 2𝑦 ′ 𝑃 − 3𝑦𝑃 = 1 + 𝑥𝑒 𝑥 [(𝐵𝑥 2 + 4𝐵𝑥 + 𝐶𝑥 + 2𝐵 + 2𝐶)𝑒 𝑥 ] + 2[𝐴 + (2𝐵𝑥 + 𝐶)𝑒 𝑥 + (𝐵𝑥 2 + 𝐶𝑥)𝑒 𝑥 ] − 3[𝐴𝑥 + (𝐵𝑥 2 + 𝐶𝑥)𝑒 𝑥 ] = 1 + 𝑥𝑒 𝑥

(𝐵𝑥 2 + 4𝐵𝑥 + 𝐶𝑥 + 2𝐵 + 2𝐶)𝑒 𝑥 + 2𝐴 + (4𝐵𝑥 + 2𝐶)𝑒 𝑥 + (2𝐵𝑥 2 + 2𝐶𝑥)𝑒 𝑥 − 3𝐴𝑥 + (3𝐵𝑥 2 + 3𝐶𝑥)𝑒 𝑥 = 1 + 𝑥𝑒 𝑥 (8𝐵𝑥 + 2𝐵 + 4𝐶)𝑒 𝑥 + 2𝐴 + 2𝐴 − 3𝐴𝑥 = 1 + 𝑥𝑒 𝑥

Calculando:  2𝐴 = 1 1 𝐴= 2  2𝐵 = 1 1 𝐵= 8  2𝐵 + 4𝐶 = 0 1 2( ) + 4𝐶 = 0 8 1 4𝐶 = − 4 1 𝐶=− 16

Remplazamos en: 𝑦𝑝 = 𝐴𝑥 + (𝐵𝑥 2 + 𝐶𝑥)𝑒 𝑥 1

1

1

 𝑦𝑝 = 2 𝑥 + (8 𝑥 2 − 16 𝑥)𝑒 𝑥

Hallamos: 𝑦𝑔 = 𝑦𝑐 + 𝑦𝑝 1

1

1

 𝑦𝑔 = 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 −2𝑥 + 2 𝑥 + 8 𝑥 2 𝑒 𝑥 − 16 𝑥𝑒 𝑥

d. 𝐲 ′′′ − 𝟕𝐲 ′ + 𝟔𝐲 = 𝟐𝐬𝐞𝐧𝐱

Hallamos yc de: 𝑦 ′′′ − 7𝑦 ′ + 6𝑦 = 0  E.C: 𝑟 2 − 7𝑟 + 6 = 0 𝑟 2 + 2𝑟 − 3 = 0 (𝑟 − 1)(𝑟 − 2)(𝑟 + 3) = 0(ruffini) 𝑟 = 1 ˅ 𝑟 = 2 ˅ 𝑟 = −3

 yc: 𝑦(𝑔) = 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 2𝑥 + 𝐶3 𝑒 3𝑥

Hallamos yp similar a 𝑓(𝑥) = 2𝑠𝑒𝑛𝑥 Caso III, 1  𝑦𝑝 = 2𝐴𝑐𝑜𝑠𝑥 + 2𝐵𝑠𝑒𝑛𝑥 Hallamos a ˄ b; hallar y''' ˄ y'' ˄ y' Si 𝑦𝑝 = 2𝐴𝑐𝑜𝑠𝑥 + 2𝐵𝑠𝑒𝑛𝑥  𝑦 ′ 𝑝 = −2𝐴𝑠𝑒𝑛𝑥 + 2𝐵𝑐𝑜𝑠𝑥  𝑦′′𝑝 = −2𝐴𝑐𝑜𝑥 − 2𝐵𝑠𝑒𝑛𝑥  𝑦′′′𝑝 = 2𝐴𝑠𝑒𝑛𝑥 − 2𝐵𝑐𝑜𝑠𝑥 Por lo tanto: 𝑦 ′′ 𝑃 −7𝑦 ′ 𝑃 + 6𝑦𝑃 = 1 + 𝑥𝑒 𝑥 2𝐴𝑠𝑒𝑛𝑥 − 2𝐵𝑐𝑜𝑠𝑥 + 14𝐴𝑠𝑒𝑚𝑥 − 14𝐵𝑐𝑜𝑠𝑥 + 12𝐴𝑐𝑜𝑠𝑥 + 12𝐵𝑠𝑒𝑛𝑥 = 2𝑠𝑒𝑛𝑥 + 0𝑐𝑜𝑥 (16𝐴 + 12𝐵)𝑠𝑒𝑛𝑥 + (12𝐴 − 16𝐵)𝑐𝑜𝑠𝑥 = 2𝑠𝑒𝑛𝑥

Calculando:  12𝐴 − 16𝐵 = 0 12𝐴 = 16𝐵 16 𝐴= 𝐵 12 4 𝐴= 𝐵 3  16𝐴 + 12𝐵 = 2 4 16 ( 𝐵) + 12𝐵 = 2 3 100𝐵 = 6 3 𝐵= 50

Reemplazando B en A

4

3

 𝐴 = 3 (50) 𝐴=

4 50

Remplazamos en: 𝑦𝑝 = 2𝐴𝑐𝑜𝑠𝑥 + 2𝐵𝑠𝑒𝑛𝑥 4

3

 𝑦𝑝 = 25 𝑐𝑜𝑠𝑥 + 25 𝑠𝑒𝑛𝑥

Hallamos: 𝑦𝑔 = 𝑦𝑐 + 𝑦𝑝 4

3

 𝑦𝑔 = 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 2𝑥 + 𝐶3 𝑒 3𝑥 + 25 𝑐𝑜𝑠𝑥 + 25 𝑠𝑒𝑛𝑥 e. 𝐲 ′′ − 𝐲 ′ = 𝐞𝐱 𝐬𝐞𝐧𝐱

Hallamos yc de 𝑦 ′′ − 𝑦 ′ = 0  E.C: 𝑟2 − 𝑟 = 0 𝑟(𝑟 − 1) = 0 𝑟=0 ˅ 𝑟=1  yc: 𝑦(𝑔) = 𝐶1 𝑒 0𝑥 + 𝐶2 𝑒 𝑥 𝑦(𝑔) = 𝐶1 + 𝐶2 𝑒 𝑥

Hallamos yp similar a 𝑓(𝑥) = 𝑒 𝑥 𝑠𝑒𝑛𝑥 Caso IV, 2  𝑦𝑝 = 𝑥𝑒

𝑥 (𝐴𝑥𝑐𝑜𝑠𝑥

+ 𝐵𝑥𝑠𝑒𝑛𝑥) 𝑦𝑝 = 𝐴𝑥 𝑒 𝑐𝑜𝑠𝑥 + 𝐵𝑥 2 𝑒 𝑥 𝑠𝑒𝑛𝑥 2 𝑥

Hallamos a ˄ b; hallar y'' ˄ y' Si 𝑦𝑝 = 𝐴𝑥 2 𝑒 𝑥 𝑐𝑜𝑠𝑥 + 𝐵𝑥 2 𝑒 𝑥 𝑠𝑒𝑛𝑥  𝑦 ′ 𝑝 = (𝐴𝑐𝑜𝑠𝑥 − 𝐴𝑥𝑠𝑒𝑛𝑥)𝑒 𝑥 + 𝐴𝑥𝑐𝑜𝑠𝑥)𝑒 𝑥 + (𝐵𝑠𝑒𝑛𝑥 + 𝐵𝑥𝑐𝑜𝑠𝑥)𝑒 𝑥 + 𝑦 ′𝑝

𝐵𝑥𝑠𝑒𝑛𝑥𝑒 𝑥 = (𝐴𝑐𝑜𝑠𝑥 − 𝐴𝑥𝑠𝑒𝑛𝑥 + 𝐴𝑥𝑐𝑜𝑠𝑥)𝑒 𝑥 + (𝐵𝑠𝑒𝑛𝑥 + 𝐵𝑥𝑐𝑜𝑠𝑥 + 𝐵𝑥𝑠𝑒𝑛𝑥)𝑒 𝑥

 𝑦′′𝑝 = (−𝐴𝑠𝑒𝑛𝑥 − 𝐴𝑠𝑒𝑛𝑥 − 𝐴𝑥𝑐𝑜𝑠𝑥)𝑒 𝑥 + (𝐴𝑐𝑜𝑠𝑥 − 𝐴𝑥𝑠𝑒𝑛𝑥)𝑒 𝑥 + (𝐴𝑐𝑜𝑠𝑥 − 𝐴𝑥𝑠𝑒𝑛𝑥) + 𝐴𝑥𝑐𝑜𝑠𝑥𝑒 𝑥 + (𝐵𝑐𝑜𝑠𝑥 + 𝐵𝑐𝑜𝑠𝑥 − 𝐵𝑥𝑠𝑒𝑛𝑥)𝑒 𝑥 + (𝐵𝑠𝑒𝑛𝑥 + 𝐵𝑥𝑐𝑜𝑠𝑥)𝑒 𝑥 + (𝐵𝑠𝑒𝑛𝑥 + 𝐵𝑥𝑐𝑜𝑠𝑥)𝑒 𝑥 + 𝐵𝑥𝑠𝑒𝑛𝑥𝑒 𝑥

 𝑦′′𝑝 = (−2𝐴𝑠𝑒𝑛𝑥 + 2𝐴𝑐𝑜𝑠𝑥 − 2𝐴𝑥𝑠𝑒𝑛𝑥 + 2𝐵𝑐𝑜𝑠𝑥 + 2𝐵𝑠𝑒𝑛𝑥 + 2𝐵𝑥𝑐𝑜𝑠𝑥)𝑒 𝑥

Por lo tanto: 𝑦 ′′ 𝑃 −𝑦 ′ 𝑃 = 𝑒 𝑥 𝑠𝑒𝑛𝑥 (−2𝐴𝑠𝑒𝑛𝑥 + 2𝐴𝑐𝑜𝑠𝑥 − 2𝐴𝑥𝑠𝑒𝑛𝑥 + 2𝐵𝑐𝑜𝑠𝑥 + 2𝐵𝑠𝑒𝑛𝑥 + 2𝐵𝑥𝑐𝑜𝑠𝑥)𝑒 𝑥 − (𝐴𝑐𝑜𝑠𝑥 − 𝐴𝑥𝑠𝑒𝑛𝑥 + 𝐴𝑥𝑐𝑜𝑠𝑥)𝑒 𝑥 + (𝐵𝑠𝑒𝑛𝑥 + 𝐵𝑥𝑐𝑜𝑠𝑥 + 𝐵𝑥𝑠𝑒𝑛𝑥)𝑒 𝑥 = 𝑒 𝑥 𝑠𝑒𝑛𝑥 (−2𝐴 − 2𝐴𝑥 + 𝐵 − 𝐵𝑥)𝑠𝑒𝑛𝑥 + (𝐴 − 𝐴𝑥 + 𝐵𝑥 + 2𝐵)𝑐𝑜𝑠𝑥 = 𝑠𝑒𝑛𝑥

Calculando:  −2𝐴 + 𝐵 = 1 𝐵 = 1 + 2𝐴  2𝐵 + 𝐴 = 0 2(1 + 2𝐴) + 𝐴 = 0 2 + 4𝐴 + 𝐴 =0 2 𝐴=− 5

Reemplazando A en B 2

 2𝐵 + (− 5)=0 𝐵=

1 5

Remplazamos en: 𝑦𝑝 = 𝐴𝑥 2 𝑒 𝑥 𝑐𝑜𝑠𝑥 + 𝐵𝑥 2 𝑒 𝑥 𝑠𝑒𝑛𝑥 2

1

 𝑦𝑝 = − 5 𝑥 2 𝑒 𝑥 𝑐𝑜𝑠𝑥 + 5 𝑥 2 𝑒 𝑥 𝑠𝑒𝑛𝑥

Hallamos 𝑦𝑔 = 𝑦𝑐 + 𝑦𝑝 2

1

 𝑦𝑔 = 𝐶1 + 𝐶2 𝑒 𝑥 − 5 𝑥 2 𝑒 𝑥 𝑐𝑜𝑠𝑥 + 5 𝑥 2 𝑒 𝑥 𝑠𝑒𝑛𝑥

5. Resolver las siguientes ecuaciones diferenciales utilizando Variación de parámetros: a. yii + y = sec(x) SOLUCION: E.C: 𝑟 2 + 1 = 0 𝑟2 − 𝑖2 = 0 ;iEe r = i v r = -i r=0+i

v

r=0–i

𝑦𝑔 = 𝐶1 𝑒 0𝑥 𝑐𝑜𝑠𝑥 + 𝐶2 𝑒 𝑜𝑥 𝑠𝑒𝑛𝑥 𝒚𝒄 = 𝒚𝒈 = 𝑪𝟏 𝒄𝒐𝒔𝒙 + 𝑪𝟐 𝒔𝒆𝒏𝒙 Luego la solución particular es: 𝑦𝑐 = 𝑢1 cos 𝑥 + 𝑢2 𝑠𝑒𝑛 𝑥 Y se establece el sistema: 𝑢1′ cos 𝑥 + 𝑢2′ 𝑠𝑒𝑛 𝑥 = 0 𝑢1′ cos 𝑥 + 𝑢2′ = sec 𝑥 Hallamos: 𝑢1 = ?

; 𝑢2 =?

0 𝑠𝑒𝑛 𝑥 ] sec 𝑥 cos 𝑥 = −𝑠𝑒𝑛𝑥. 𝑠𝑒𝑐𝑥 𝑢1 = 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥 𝑐𝑜𝑠 2 𝑥 + 𝑠𝑒𝑛2 𝑥 [ ] −𝑠𝑒𝑛𝑥 𝑐𝑜𝑠𝑥 [

1 𝑢1′ = −𝑠𝑒𝑛 𝑥. ( ) = −𝑡𝑔 𝑥 cos 𝑥 𝑢1′ = − ∫ 𝑡𝑔 𝑥 𝑑𝑥 𝒖𝟏 = 𝑳𝒏(𝐜𝐨𝐬 𝒙) Luego:

𝑢2′

[ =

𝑐𝑜𝑠𝑥 0 ] 𝑐𝑜𝑠𝑥. 𝑠𝑒𝑐𝑥 −𝑠𝑒𝑛𝑥 secx 𝑥 = 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥 𝑐𝑜𝑠 2 𝑥 + 𝑠𝑒𝑛2 𝑥 [ ] −𝑠𝑒𝑛𝑥 𝑐𝑜𝑠𝑥

𝑢2′ = cos 𝑥 . 𝑠𝑒𝑐𝑥 1

𝑢2′ = cos 𝑥 . (cos 𝑥) = 1 Integramos: 𝑢2 = ∫ 𝑑𝑥 = 𝑥 Reemplazamos en la solución particular 𝑦𝑝 = [𝐿𝑛(𝑐𝑜𝑠𝑥)]𝑐𝑜𝑠𝑥 + 𝑥𝑠𝑒𝑛 𝑥 𝑦𝑔 = 𝑦𝑐 + 𝑦𝑝

Finalmente:

𝑌𝑔 = 𝐶1 𝑐𝑜𝑠𝑥 + 𝐶2 𝑠𝑒𝑛𝑥 + [𝐿𝑛(𝑐𝑜𝑠𝑥)]𝑐𝑜𝑠𝑥 + 𝑥𝑠𝑒𝑛𝑥 b. yii + 4y = 4ctg2x SOLUCION: E.C:

𝑟2 + 4 = 0 r = 2i

v r = -2i

𝒚𝒄 = 𝒚𝒈 = 𝑪𝟏 𝒄𝒐𝒔𝟐𝒙 + 𝑪𝟐 𝒔𝒆𝒏𝟐𝒙 Luego la solución particular es: 𝑦𝑝 = 𝑢1 cos 2𝑥 + 𝑢2 𝑠𝑒𝑛 2𝑥 Y se establece el sistema: 𝑢1′ cos 2𝑥 + 𝑢2′ 𝑠𝑒𝑛 2𝑥 = 0 −2𝑢1′ sen 2𝑥 + 2𝑢2′ 𝑐𝑜𝑠2𝑥 = 4 𝑐𝑡𝑔 2𝑥 Hallamos: 𝑢1 = ?

; 𝑢2 =?

0 𝑠𝑒𝑛 2𝑥 ] −𝑠𝑒𝑛2𝑥. 4𝑐𝑡𝑔 2𝑥 4𝑐𝑡𝑔2𝑥 2cos 2𝑥 𝑢1 = = 𝑐𝑜𝑠2𝑥 𝑠𝑒𝑛2𝑥 2𝑐𝑜𝑠 2 𝑥 + 2𝑠𝑒𝑛2 𝑥 [ ] −2𝑠𝑒𝑛𝑥 2𝑐𝑜𝑠2𝑥 [

𝑢1′ = −2𝑐𝑜𝑠 2𝑥 1

𝑢1 = −2 ∫ 𝑐𝑜𝑠 2𝑥 𝑑𝑥 = −2. 2 𝑠𝑒𝑛2𝑥 𝒖𝟏 = −𝒔𝒆𝒏 𝟐𝒙 Luego: 𝑐𝑜𝑠2𝑥 0 ] 𝑐𝑜𝑠2𝑥. 4𝑐𝑡𝑔2𝑥 −2𝑠𝑒𝑛2𝑥 4ctg 2𝑥 ′ 𝑢2 = = 𝑐𝑜𝑠2𝑥 𝑠𝑒𝑛2𝑥 2𝑐𝑜𝑠 2 𝑥 + 2𝑠𝑒𝑛2 𝑥 [ ] −2𝑠𝑒𝑛2𝑥 2𝑐𝑜𝑠2𝑥 [

𝑢2′ = 2 cos 2𝑥 . 𝑐𝑡𝑔 2𝑥 𝒖′𝟐 = 𝟐𝒄𝒐𝒔𝟐𝒙. 𝒕𝒈 𝟐𝒙 Integramos: 1 1 𝑢2 = 2 ∫ 𝑐𝑜𝑠2𝑥. 𝑡𝑔 2𝑥𝑑𝑥 = 𝑐𝑜𝑠2𝑥 − 𝐿𝑛(2𝑥 + 1) + 𝐿𝑛(1 − cos 2𝑥) 2 2 1 1 − cos 2𝑥 𝑢2 = = 𝑐𝑜𝑠2𝑥 + 𝐿𝑛( ) 2 1 + cos 2𝑥 Reemplazamos en la solución particular 1 1 − cos 2𝑥 𝑦𝑝 = [𝑠𝑒𝑛 2𝑥]𝑐𝑜𝑠2𝑥 + [ 𝑐𝑜𝑠2𝑥 + 𝐿𝑛( ) ]𝑠𝑒𝑛 2𝑥 2 1 + cos 2𝑥 Finalmente:

𝑦𝑔 = 𝑦𝑐 + 𝑦𝑝

𝑌𝑔 = 𝐶1 𝑐𝑜𝑠2𝑥 + 𝐶2 𝑠𝑒𝑛2𝑥 + [−𝑠𝑒𝑛 2𝑥]𝑐𝑜𝑠2𝑥 + [ 𝑐𝑜𝑠2𝑥 1 1 − cos 2𝑥 + 𝐿𝑛( ) ]𝑠𝑒𝑛 2𝑥 2 1 + cos 2𝑥

c. yii + 2yi + 2y = e-xsecx SOLUCION: 𝑟 2 + 2𝑟 + 2 = 0

E.C:

r = -1 + i

v

r = -2i

𝒚𝒄 = 𝒚𝒈 = 𝑪𝟏 𝒆−𝒙 𝒄𝒐𝒔𝟐𝒙 + 𝑪𝟐 𝒆−𝒙 𝒔𝒆𝒏𝟐𝒙 Luego la solución particular es: 𝑦𝑝 = 𝑢1 𝑒 −𝑥 𝑐𝑜𝑠𝑥 + 𝑢2 𝑒 −𝑥 𝑠𝑒𝑛𝑥 Y se establece el sistema: 𝑢1′ 𝑒 −𝑥 cos 𝑥 + 𝑢2′ 𝑒 −𝑥 𝑠𝑒𝑛 𝑥 = 0 −𝑢1′ 𝑒 −𝑥 sen 𝑥 + 𝑢2′ 𝑒 −𝑥 𝑐𝑜𝑠𝑥 = 𝑒 −𝑥 𝑠𝑒𝑐𝑥 Hallamos: 𝑢1 = ?

𝑢1′ =

[

0 𝑒 −𝑥 𝑠𝑒𝑐𝑥 𝑒 −𝑥 𝑐𝑜𝑠𝑥

[ −𝑥 −𝑒 𝑠𝑒𝑛𝑥

; 𝑢2 =? 𝑒 −𝑥 𝑠𝑒𝑛𝑥 −2𝑥 ] 𝑒 −𝑥 𝑐𝑜𝑠𝑥 = 𝑒 𝑠𝑒𝑛𝑥. 𝑠𝑒𝑐𝑥 𝑒 −𝑥 𝑠𝑒𝑛𝑥 𝑒 −2𝑥 ] −𝑥 𝑒 𝑐𝑜𝑠𝑥

𝑢1′ = −𝑡𝑔 𝑥 𝒖𝟏 = − ∫ 𝒕𝒈 𝒙 𝒅𝒙 = 𝑳𝒏 (𝒄𝒐𝒔 𝒙) Luego:

𝑢2′ =

[

𝑒 −𝑥 𝑐𝑜𝑠𝑥 0 −𝑥 ] −𝑒 −𝑥 𝑠𝑒𝑛𝑥 𝑒 −𝑥 sec 𝑥 = 𝑒 𝑐𝑜𝑠𝑥. sec 𝑥 𝑒 −𝑥 𝑐𝑜𝑠2𝑥 𝑠𝑒𝑛𝑥 𝑒 −2𝑥 [ −𝑥 ] −𝑒 𝑠𝑒𝑛𝑥 𝑐𝑜𝑠𝑥

𝑢2′ = 1 Integramos: 𝑢2 = ∫ 𝑑𝑥 = 𝑥

𝑢2 = = 𝑥 Reemplazamos en la solución particular 𝑦𝑝 = [𝐿𝑛(cos 𝑥)]𝑒 −𝑥 𝑐𝑜𝑠𝑥 + 𝑒 −𝑥 𝑠𝑒𝑛𝑥 Finalmente:

𝑦𝑔 = 𝑦𝑐 + 𝑦𝑝

𝑌𝑔 = 𝐶1 𝑒 −𝑥 𝑐𝑜𝑠𝑥 + 𝐶2 𝑒 −𝑥 𝑠𝑒𝑛𝑥 + [𝐿𝑛(cos 𝑥)]𝑒 −𝑥 𝑐𝑜𝑠𝑥 + 𝑒 −𝑥 𝑠𝑒𝑛𝑥

d. yii – 3yi + 2y =

e2x 1+ e2x

SOLUCION:

∗) 𝑟 2 − 3𝑟 + 2 = 0 (𝑟 − 1)(𝑟 − 2) = 0 𝑟=1

𝑣

𝑟=2

𝒀𝒈 = 𝑪𝟏 𝒆𝒙 + 𝑪𝟐 𝒆𝟐𝒙 ∗∗) 𝑦𝑔 = 𝑢1 𝑒 𝑥 + 𝑢. 𝑒 2𝑥

𝑢1′ 𝑒 𝑥 + 𝑢2′ 𝑒 2𝑥 = 0 𝑢1′ 𝑒 𝑥 + 𝑢2′ 𝑒 2𝑥 =

[ 𝑢1′ =

𝑢1′ =

𝑒 2𝑥 1 + 𝑒 2𝑥

0 𝑒 2𝑥 ] 𝑒 2𝑥 2𝑒 2𝑥 2𝑥 1+ 𝑒 𝑒 𝑥 𝑒 2𝑥 [ 𝑥 ] 𝑒 2𝑒 2𝑥 𝑒𝑥 1 + 𝑒 2𝑥 2𝑒 3𝑥 . 𝑒 3𝑥

−𝑒 2𝑥 .

𝑢1′ =

𝑒𝑥 1 + 𝑒 2𝑥 𝑒 3𝑥

−𝑒 3𝑥 .

𝑢1′ =

𝑒𝑥 1 + 𝑒 2𝑥

𝑢1′ = − ∫ 𝑢1 = − ∫

𝑒𝑥 𝑑𝑥 1 + 𝑒 2𝑥

𝑑(𝑒 𝑥 ) 𝑒𝑥 = −𝑎𝑟𝑐𝑡𝑔 [ ] 1 + (𝑒 𝑥 )2 1

𝒖𝟏 = −𝒂𝒓𝒄𝒕𝒈[𝒆𝒙 ] Luego: 𝑒𝑥

[ 𝑢2′ =

𝑢2′ =

𝑢2′

𝑒 3𝑥 .

0 2𝑥 ] 𝑒 𝑒𝑥 2𝑥 1+ 𝑒 𝑒 𝑥 𝑒 2𝑥 [ 𝑥 ] 𝑒 2𝑒 2𝑥

𝑒𝑥 1 + 𝑒 2𝑥 𝑒 3𝑥

=

1 1 + 𝑒 2𝑥

1 1 2𝑥 𝑒 = ∫ 𝑑𝑥 = ∫ 𝑑𝑥 1 + 𝑒 2𝑥 1 + 𝑒 2𝑥 𝑒 2𝑥 𝑢2 = ∫

𝑒 −2𝑥 𝑑𝑥 1 + 𝑒 −2𝑥

−1 𝑑(1 + 𝑒 −2𝑥 ) 𝑢2 = ∫ 2 1 + 𝑒 −2𝑥 𝒖𝟐 =

−𝟏 𝑳𝒏(𝟏 + 𝒆−𝟐𝒙 ) 𝟐

Luego: 𝑦𝑝 = 𝑢1 𝑒 𝑥 + 𝑢2 𝑒 2𝑥 1 𝑦𝑝 = −𝑎𝑟𝑐𝑡𝑔(𝑒 𝑥 )𝑒 𝑥 − 𝐿𝑛(𝟏 + 𝒆−𝟐𝒙 ) 2

𝑦𝑔 = 𝑦𝑐 + 𝑦𝑝

Finalmente:

𝟏 𝒚𝒈 = 𝑪𝟏 𝒆𝒙 + 𝑪𝟐 𝒆𝟐𝒙 − 𝒆𝒙 𝒂𝒓𝒄𝒕𝒈(𝒆𝒙 ) − 𝒆𝟐𝒙 𝑳𝒏(𝟏 + 𝒆−𝟐𝒙 ) 𝟐

e. yii + y = xcosx

SOLUCION: E.C: 𝑟 2 + 1 = 0 𝑟2 − 𝑖2 = 0 ;iEe r = i v r = -i r=0+i

v

r=0–i

𝑦𝑔 = 𝐶1 𝑒 0𝑥 𝑐𝑜𝑠𝑥 + 𝐶2 𝑒 𝑜𝑥 𝑠𝑒𝑛𝑥 𝒚𝒄 = 𝒚𝒈 = 𝑪𝟏 𝒄𝒐𝒔𝒙 + 𝑪𝟐 𝒔𝒆𝒏𝒙 Luego la solución particular es: 𝑦𝑐 = 𝑢1 cos 𝑥 + 𝑢2 𝑠𝑒𝑛 𝑥 Y se establece el sistema: 𝑢1′ cos 𝑥 + 𝑢2′ 𝑠𝑒𝑛 𝑥 = 0 𝑢1′ cos 𝑥 + 𝑢2′ = xcos 𝑥 Hallamos: 𝑢1 = ?

𝑢1′

; 𝑢2 =?

0 𝑠𝑒𝑛 𝑥 ] −𝑠𝑒𝑛𝑥. 𝑥𝑐𝑜𝑠𝑥 xcos 𝑥 cos 𝑥 = = 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥 𝑐𝑜𝑠 2 𝑥 + 𝑠𝑒𝑛2 𝑥 [ ] −𝑠𝑒𝑛𝑥 𝑐𝑜𝑠𝑥 [

𝑢1′ = − ∫ 𝑥𝑠𝑒𝑛𝑥. 𝑐𝑜𝑠𝑥 𝑑𝑥 =

1 1 𝑥𝑐𝑜𝑠(2𝑥) − 𝑠𝑒𝑛(2𝑥) 4 8

𝑢2′ =

[

𝑐𝑜𝑠𝑥 0 ] 𝑥𝑐𝑜𝑠 2 𝑥 −𝑠𝑒𝑛𝑥 𝑥𝑐𝑜𝑠𝑥 = 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥 𝑐𝑜𝑠 2 𝑥 + 𝑠𝑒𝑛2 𝑥 [ ] −𝑠𝑒𝑛𝑥 𝑐𝑜𝑠𝑥

𝑢2 = ∫ 𝑥𝑐𝑜𝑠 2 𝑥𝑑𝑥 = Finalmente:

1 2 1 1 𝑥 + 𝑥𝑠𝑒𝑛(2𝑥) − cos(2𝑥) 4 4 8

𝑦𝑔 = 𝑦𝑐 + 𝑦𝑝

1 1 1 𝑦𝑔 = 𝐶1 𝑐𝑜𝑠𝑥 + 𝐶2 𝑠𝑒𝑛𝑥 + [ 𝑥𝑐𝑜𝑠(2𝑥) − 𝑠𝑒𝑛(2𝑥)] 𝑐𝑜𝑠𝑥 − cos(2𝑥) 4 8 8 1 2 1 1 + [ 𝑥 + 𝑥𝑠𝑒𝑛(2𝑥) − cos(2𝑥)] 𝑠𝑒𝑛𝑥 4 4 8