Ejercicio 3.1.2 Ecuaciones homogéneas Del ejercicio 15 a 22 determine determ ine si el conjunto de funciones es dado es linea lmente independiente en el intervalo (- infinito a infinito).
15. F1(x) = x, f 2(x) = x2, f 3(x) = 4x-3x2 W(x, x2, 4x-3x2) = x
x2
4x- 3x2
1
2x
4- 6x
0
2
-6
X (-6x – (-6x – (8-12x)) (8-12x)) – X2 (-6-0) + 4x-3x2 (2-0) – X 2 2 2 X (-8+6x) – (-8+6x) – x x2 (-6) + 4x-3x2(2) = -8x +6x2 – 6x – 6x +8x-6x = -6x L.I
16. f 1(x)=0, f 2(x)= x, f 3(x)= ex W (0, x, e x)= 0
x
0
1
0
0
ex e
x
e
x
0(ex – 0) – 0) – – x x (0-0) + e x (0-0) = 0 (e x) –x –x (0)+ e x (0) = 0 L.D
17. F1(x) = 5, f 2(x) = x2, f 3(x) = sen2 x W (5, x2, sen2 x) = 5
x2
0
2x
0
2
sen2x 2cosx -2senx
2
2
5 (-4xsenx – (-4xsenx – 4cosx) 4cosx) – – x x (0-0) + sen x (0-0) = -20x senx – senx – 20cosx 20cosx
18. f 1(x)= cos2x, f 2(x)= 1, f 3(x)= cos2x W (cos2x, 1, cos2x) = cos2x
1
cos2x
-2sen2x
0
-2senx
-4cos2x
0
-2cosx
L.I
cos2x (0-0) -1(4sen2xcosx – 8cos2xsenx) + cos2x (0-0) = -4sen2xcosx + 8cos2xsenx
L.I
19. F1(x) =x, f 2(x) = x-1, f 3(x) = x+3 W(x, x-1, x+3) = x
x-1
x+3
1
1
1
0
0
0
X (0) - x-1 (0) + x+3 (0) = 0 L.D
20. f 1(x)= 2+x, f 2(x)= 2 + x W(2+x, 2+-x)=
2+x 1 0
2-x
2+x
-1
1
0
0
2+x (0-0) – 2-x(0-0) + 2+x(0-0)= 0
L.D
21. F1(x) =1+x, f 2(x)=x, f 3(x)= x2 W(1+x, x ,x2)= 1+x 1 0
x
x2
1
2x
0
2
1+x(2-0) –x(2-0) + X2(0) = 2+2x-2x= 2
x
-x
22. f 1(x)= e , f 2(x)= e , f 3(x)= sen x W(ex, e-x, senx)= ex ex
e-x
senx
-e-x
cosx
L.I
ex
e-x
-senx
ex (ex senx - e-x cosx) - e-x ( - ex senx - e xcosx) + senx (e 0-e0) = e2x senx – e0 cosx + e 0 senx + e 0cosx= e2x senx + senx 4. y’’ -3y’ + 2y=0
L.I
Ejercicio 3.3 En los problemas 1 a 14, encuentre la solución general de la ecuación diferencial de segundo orden dado. 1.
4y’’ + y’
4m2 +m m(4m +1) m1=0 m2= -1/4 Y= C1 e0x + C2 e -1/4x
2.
y’’ – 36y=0
m2 – 36=0 (m + 6) (m + 6) m1=-6
m 2=6
Y= C1 e-6x + C2 e6x
3.
y’’ – y’ -6y=0
m2 –m -6=0 (m +2) (m -3) m1=-2 m2=3 Y= C1e-2x + C2e3x
4. 2y=0
y’’ -3y’+
m2 – 3m +2=0 (m -2) (m-1) m1=2 m2=1 Y= C1e2x + C2 ex
5.
y’’ + 8y’ +16y=0
m2 + 8m +16=0 (m +4) (m +4) m1,2= -4 Y= C1e-4x + C2xe-4x
6.
y’’ -10y’ +25y=0
m2 -10m +25=0 (m -5) (m -5) m1,2= 5 Y= C1e5x +C2xe5x
7.
12y’’ – 5y’ -2y=0 2
12m – 5m -2=0 m1,2= -(-5) +-
(-5)2 -4(12)(-2) / 2(12) = 5 +- 11/ 24
m1= 2/3 m 2= - 1/4 Y= C1 e2/3x + C2 e -1/4X 8.
y’’ + 4y’ –y=0
m2 +4m-1=0 m1,2= -4 +- (4)2 -4(1)(-1) / 2(1) = -4 +- 4(2)1/2/2 = -2 +- 2(2)1/2 y= C1 e -2 +2(2)1/2 x + C2 e -2- 2(2)1/2 x
9.
y’’ +9y=0
m2 +9=0 m2= -9
m= (=9)1/2
m1= 3i m 2= -3i
Y= C1 cos 3x + C2 sen3x
10.
3y’’+ y +=0
3m2 +1=0 m1= 1/ (3)1/2
m2= - 1/(3)1/2
Y= C1 cos x/ (3)1/2 + C2 sen x/(3)1/2
11.
y’’ -4y’ +5y=0
m2 – 4m +5y=0 m1,2= -(4) +- (-4)2 -4(-1)(5) / 2(1) = -4 +- (-4)1/2 / 2= 2 =- 2i Y= C1 e2x cosx + C2 e2x senx
12.
2y’’ + 2y’ + y=0
2m2 +2m +1=0 m1,2= -2 +-
(2)2 – 4(2)(1) / 2(2) = -2 +- 2i/4= - ½ +- ½ i
Y= e -1/2x (C1 cos x/2 + C2 senx/2)
13.
3y’’ +2y’ + y=0
3m2 +2m +1+0 m1,2= -2 +-
22- 4(3)(1)/ 2(3) = -2 +- 2 (2) 1/2 i / 6 = -1/3 +- 1/3 (2) 1/2 i
Y= e -1/3x (C1 cos (2)1/2/x + C2 sen (2)1/2/x)
14.
2y’’ -3y’ +4y=0
2m2 -3m +4m=0 (-3)2 -4(2)(4) / 2(2) = ¾ + (23) 1/2/4 i
m1,2= -(-3) +-
Y= e3/4x (C1 cos (23)1/2 / 4 x +C2 sen (23) 1/2 / 4 x)
Ejercicios 3.4 En los problemas 1 a 26, resuelva la ecuación diferencial dada mediante coeficientes indeterminados. 1. y” +3y’+2y=6
2+3
+2=0
m1=-1 m2=-2 yc=c1 − + 2 −2 2A=6 A=3 yp= 3 y=c1 − + 2 −2 +3 2. 4y” +9y=15 4 2+9=0 m1=−3 2 I
m2=−3 2 I yc=c1cos3 2 x+c2sen−3 2 x
yp=A 9A=15 A=5/3 YP=5/3 Y=c1cos3 2x+c2sen 3 2 +5 3
3. y”-10y’+25y=30x+3
2 + 10
+ 25 = 0
yc=c1 5 + 2
5 +6/5x+6/5
25A=30, -10A+25B=3 A=6/5 B=6/5 m1=m2=5 yc=c1 5 + 2
5 +6/5x+6/5I
4. y” +y’-6y=2x
2+
− 6 =0
m1=-3 m2=2 yc=c1 −3 + 2 2 yp=Ax+B -6A=2 A-6-6B=0 A=−1/3; B=− 1/18 yp=−1/3 - 1/18 yc=c1 −3 + 2 2 – 1/3
−1/18
5. 1 4
" − 8 ′ +
= 2 − 2
14 2+
+1=0
yc=c1 −2 + 2 yp=
2+
−2
+
A=1 2A+B=-2 1/2A+B+C A=1 B=-4 C=7 2 Y= c1 −2 + 2
−2 + 2-4x+7/2
6. y”-8y’+20y=
2−8
-26x
+ 20 = 0
m1=3+4i m2=2-4i Y= 2 ( 1
4 + 2
yp=
+
2+
4 )
2A-8B+20C=0 -6D+13E=0 -16A+20B=0 13D=-26 20A=100 A=5, B=4, C=11/10,
= −2, E=-12/13
yp= 5 2 + 4 + 11 10 + (−2 −12 13) y= 2 ( 1
4 + 2
4 ) + 5 + 4 + 11/10 + (−2 −12/13)
7. y” +y’-6y=2x
2+3=0 m1=√3 ; m2=-√3 yc=c1cos√3 + 2 yp= (
2+
√3
+ ) 3
12A=-48 A=-4 B=4 C=-4/3; YP= (-4 2 + 4 y= c1cos√3 + 2
− 4/3)
√3 +(−4 2 + 4 − 4/3) 3
8. 4y”-4y’-3y=cos2x
2+
− 6 =0
m1=-3 m2=2 yc=c1 −3 + 2 2 yp=Ax+B -6A=2 A-6-6B=0 A=−1/3; B=− 1/18 yp=−1/3 - 1/18 yc=c1 −3 + 2 2 – 1/3 −1/18
9. y”-y’=-3
2 + 10
yc=c1 5 + 2
+ 25 = 0
5 +6/5x+6/5
25A=30, -10A+25B=3 A=6/5 B=6/5 m1=m2=5 yc=c1 5 + 2
5 +6/5x+6/5I
10. Y” +2y’=2x+5 - −
m 2 +2m=0 ∴ m 1 =1 y m 2 =0; y c = C 1 y p = Ax 2 +Bx+Cx −2
+C 2
2A+2B=5, 4A=2, - 2C= - 1 =1/2; = 2; =1/2, 2 x+5 = y''(x)+y'(x)+ −2 ∴
=1/2 2 + 2 + 1/2 x −2
y''(x) = - y'(x)+2 x −2 +5 y''(x)+y'(x) = −2 (2 2 x+5 2 - 1) ∴
y = C1 −2 +C2+ 1/2 x2+2 x 1/2 x −2
11. Y”- y’ +
=
+
/
m 2 - m+ 1/4 =0 ∴ m 1 =m 2 = 1/2; Yc= C1 /2 +C2x /2; y p = A+Bx 2
/2;
1/4 A= 3; 2B=1 ∴
= 12;
=1/2;
=1/2 2
4 (- y''(x)+y'(x)+
/2 +3) = y(x)
y''(x) = y'(x) − ( ) 4 +
/2 +3
y''(x)+ 1/4 (y(x) - 4 y'(x)) = ∴
y= C 2
/2 x+C 1
12. y”-16y=2
/2
/2 +3
/2 +1/2
/2 x 2 +12
m 2 - 16= 0; m1= 4; m2= - 4; yc= C 1 4 +C 2 x −4 , Yp= Ax 4
8A=2 =1/4; =1/4 4 y''(x) = 2 (8 y(x)+ 4 ) ∴
y''(x) = 16 y(x)+2 4 ∴
y= C 1 4 +C 2 −4 + 1 4 x 4
13. y” +4y= 3 sen2x
m 2 +4=0; m1=2i; m2= - 2i; Yc= c 1 cos2x+c 2 sen2x Yp= Axcos2x+Bxsen2x 4B=0; - 4A=3 ∴
= −3/4;
= 0;
= −3/4
2
y''(x)+4 y(x) = 6 Sen(x) Cos(x) y''(x) = 3 Sen(2 x) - 4 y(x) ∴
y= C 2 Sen(2x) +C 1 Cos(2x) – 3/4
14. y” +4y=(x2-3) sen2x
2
m 2 +4=0; m1= 2i; m2= - 2i; Yc= c 1 cos2x+c 2 sen2x Yp= (Ax 3 +Bx 2 +Cx) cos2x + (D x 3 +E x 2 +F x) Sen2x 2B +4F=0 6ª+8E=0 12D=0 - 4C+2E= - 3 - 8B+6D=0 - 12A= - 1 ∴
= −1/12;
= 0;
=25/32;
= (−12 3 +25/32)
= 0;
=1/16;
2 +1/16 2
= 0;
2
(x 2 - 3) sen(2 x)+4 y(x) = y''(x) y''(x) - 4 y(x) = 2 (x 2 - 3) sin(x) cos(x) y''(x) - 4 y(x) = 2 x 2 sin(x) cos(x) - 6 sin(x) cos(x) ∴
y = C 1 cos2x+C 2 sen2x + (1/12 x 3 +25/32 x) cos2x+ 1/16 x 2 sen2x
15. y”+y= 2xsenx m 2 +1 =0; m1= i; m2= - i; Yc= c 1 cos x+c 2 sen x
Yp= (Ax 2 +Bx)cosx+(Cx 2 +Dx)senx - 4A=2; - 2B+2C=0 ∴
A= 1 2; B=0; C=0; D= 1/2; Yp= 1/2 2
+ 1/2
y''(x) = 2 x sin(x) - y(x) ∴
y = C 2 senx+ C 1 cosx 1/2 x 2 cosx+1/2 x senx
16. y”-5y’=2x3-4x2-x+6 m 2 - 5m=0; m1=5; m2=0; Yc= C 1 5 +C 2
Yp= Ax 4 + Bx 3 +Cx 2 +Dx - 20A=2; 12A - 15B= - 4; 6B - 10C= - 1; 2C - 5D=6 ∴
= −1/10;
=14/75;
=63/250;
= −697/625;
= −1/10 3 +63/250 2 −29/625
5 y'(x)+x (2 (x - 2) x - 1) +6 = y''(x) y''(x) = 2 x 3 - 4 x 2 +5 y'(x) - x+6 ∴
= 1 5 + 2 − 4/10+14 3/75−53 2/250−697 /625
17. y”-2y’+5y=
yp=Ax
cos2x 2 - 2m+5=0; m1=1+2i; m2=1−2i; yc=
cos2x+Bx
(c 1 cos 2x+c 2 sin 2x)
sin 2x
4B = 1 − 4A = 0.
∴
= 0,
= 4,
=1/4
y=
(c 1 cos 2x + c 2 sin 2x) + 1/4 x
18. y”-2y’+2y=
2 , sin 2x.
(cosx-3senx) m 2 - 2m+2=0; m1= 1+i; m2 =1−i.;
(c 1 cosx+c 2 sinx)
yp = A 2 cos x+B 2 senx A+2B = 1 − 2A+B = −3.
A = 7/5, B = −1/5, yp = 7 / 5 2 cos – 1 /5 2 sin x ∴
=
( 1
+ 2
) +7/5 2
−1/5 2
19. y” +2y’+y=senx+3cos2x m 2 +2m+1=0; m1=m2= −1;
yc =C 1
+c 2 x
yp = Acos x+Bsin x+Ccos 2x+Dsin2x 2B = 0, −2A = 1, − 3C +4D = 3, − 4C − 3D = 0.
A = −12, B = 0, C = −9/25, D = 12/25, yp = − 1/2 cos x − 9/25 cos 2x + 12/25 sin 2x ∴
= 1 − + 2
− −1/2
−9/25
2 +12/25
2
20. y” +2y’-24y=16-(x+2) − m 2 + 2m −24 = 0; m1=−6; m2= 4.
yc = c 1 −6 + c 2 −4 yp =A +(Bx 2 + Cx) −4 − 24A = 16,
2B + 10C = −2, 20B = −1
∴
∴
= −2/3,
= −1/20,
= −19/100,
= −2/3− (1/20 2 +19/100 ) 4
= 1 −6 + 2 4 −2/3−1/20 2 +19/100
4
21. y”’-6y” =3-cosx
m 3 − 6m 2 = 0; m1=m2=0; m3 = 6. yc = c 1 + c 2 x + c 3 6 yp = Ax 2 +Bcosx+Csin x. − 12A = 3,
6B − C = −1, B + 6C = 0. A = − 1/4, B = −6/37, C = 1/37, yp = − 1/4 x2 − 6/37 cos x+ 1/37 sin x, ∴
y = c 1 + c 2 x + c 3 6 – 1/4 x 2 −6/37 cos x +1/37 sin x.
22. Y”’ -2y”-4y’+8y=6x
m 3 − 2m 2 − 4m+8 = 0; m1=m2=2; m3= - 2 yc = c 1 2 + c 2 x 2 + c 3 −2 yp = (Ax 3 + Bx 2) 2 24A = 6 6A + 8B = 0. A= 1/4, B = − 3/16, yp = (1/4 x3 − 3/16 x 2) e 2x, ∴
= 1 2 + 2
2 + 3 −2 + (1/4 3 −3/16 2) 2 .
23. y”’-3y” +3y’-y= x−
Yc= c 1
+c2x
yp = Ax+B+Cx 3
m 3 − m 2 −4m+4 = 0; m1=1; m2= 2 ’ m3 =−2.
+c3x2 .
− A = 1,
3A−B = 0 6C = −4. A = −1, B = −3, C = − 2 3, yp = −x−3− 2 3 x 3
∴
y=c1
+c2x
24. y”’-y”-4y’+4y= 5−
yc = c 1
+c3x2
+
−x−3− 2 3 x 3
m 4 +2m 2 +1 = 0; m1=m3=I; m2=m4=−i.
+ c 2 2 + c 3 −2
yp = A + Bx
+ Cx 2
4A = 5, − 3B = −1,
4C = 1. A = 5/4, B = 1/3, C = 1/4, yp = 5 4 + 1/3 x ∴
y= c 1
+ c 2 2 + c 3 −2 + 5 4 + 1 3 x
+ 1/4 x 2 +14x 2
25. Y (4) +2y” +y=(x-1)2 m 4 +2m 2 +1 = 0; m1=m3=I; m2=m4 =−i.
yc = c 1 cos x+c 2 sin x+c 3 x cos x+c 4 x sin x yp = Ax 2 + Bx + C. A= 1, B = −2, 4A + C = 1. A=1, B =−2, C= −3, yp = x 2 − 2x − 3, ∴
= 1
+ 2
+ 3
+ 4
+ 2 − 2 − 3.
26. Y4-y” =4x+2x −
m 4 − m 2 = 0; m1=m2=0; m3=1; m4 = −1. yc = c 1 + c 2 x + c 3
+c4 −
yp = Ax 3 + Bx 2 + (Cx 2 + Dx) − − 6A = 4, − 2B = 0,
10C−2D = 0, − 4C = 2.
A = − 2/3, B = 0, C = − 1/2, D = − 5/2, yp = − 2/3 x 3 – 1/2 x 2 + 5/2 x − ∴
y=c1+c2x+c3
+ c 4 − – 2/3 x 3 – (1/2 x 2 + 5/2 x) − .
En los siguientes ejercicios resuelva el problema de valor inicial dado. 27. y” +4y=-2, y (
) = / , y’ (
) =2
yc = c 1 cos 2x+c 2 sin 2x yp = A. A = − 1/2 ∴
y = c 1 cos2x + c 2 sin 2x−. ½
28. 2y” +3y’-2y=14x2-4x-11; y (0) =0, y’ (0) =0
yc = c1 −
+c2
/
yp = Ax2 +Bx+C. A = −7, B = −19, C = −37. y = c1 −
+ c2
/ − 7x2 − 19x – 37
Condición inicial c1 = −1/5 y c2 = 186/5 y = − 1/5 −
∴
+ 186/5
/ − 7x 2 − 19x − 37.
29. 5y” +y=-6x, y (0) =0, y ’ (0) =-10
yc = c 1 − / + c 2 yp = Ax 2 + Bx A = −3 y B = 30. y = c 1 − / + c 2 − 3x 2 + 30x. Condicion inicial. c1 = 200 y c2 = −200, y = 200 − x/5 − 200 − 3x 2 + 30x.
∴
30. y”-y=coshx, y (0) =2, y’ (0) =12
yc = c 1 cosh x + c 2 sinh x yp = Ax cosh x + Bxsinh x. A = 0 y B = 1/2 y = c 1 cosh x + c 2 sinh x + 1/2 x sinh x. Condicion inicial c 1 =2 y c 2 =12 ∴
y = 2 cosh x + 12 sinh x + 1/2 x sinh x.
En los ejercicios siguientes, resuelva el problema de valores en la frontera dado.
31. y”+y= x2+1, y(0)=5, y(1)=0
yc = c 1 cos x+c 2 sin x yp = A 2 + Bx + C. A = 1, B = 0, y C= - 1 y= c 1 cos x+c 2 sin x+x 2 −1. y (0) = 5 y y (1) = 0 Se obtiene c1 − 1 = 5 (cos1) c 1 + sin (1) c 2 = 0. y = 6 cos x − 6(cot 1) sin x + x 2 − 1.
∴
32. Y” -2y’ +2y=2x-2, y (0) =0, y( )=
yc = e x (c 1 cos x + c 2 sin x) yp = Ax + B. A = 1 y B = 0. y = e x (c 1 cos x + c 2 sin x) + x. y (0) = 0 y y(π) = π Se obtiene c1 = 0 π − eπc 1 = π. ∴
y = c2e x sin x + x.
33. y” +3y=6x, y (0) =0, y (1) +y’ (1) =0
yc = c 1 cos 2x + c 2 sin 2x yp = Acos x + B sin x en [0, π/2]. A = 0 y B = 1/3. y = c 1 cos 2x+c 2 sin 2x+ 1/3 sin x en [0, π/2]. En (π/2, ∞) y = c 3 cos 2x + c 4 sin 2x y (0) = 1 y y (0) = 2 Se obtiene c1 = 1 y 1/3+ 2c 2 = 2. y(x)= cos 2x + 56 sin 2x + 13 sin x, 0 ≤ x ≤ π/2 y(x)= 2/3 cos 2x + 5/6 sin 2x, x > π/2
