книги / Математика. Дифференциальные уравнения
.pdf
11.16.а) y′′+ 2xy′2 = 0,
11.17.а) 2xy′y′′= y′2 +1,
11.18.а) y′′− xy−′1 = x(x −1),
11.19.а) y′′′+ y′′tgx =sec x,
11.20.а) y′′−2 y′c tgx =sin3 x,
11.21.а) y′′+ 4 y′= 2x2 ,
11.22.а) xy′′− y′= 2x2ex ,
11.23.а) x( y′′+1) + y′= 0,
11.24.а) y′′+ 4 y′ = cos 2x,
11.25.а) y′′+ y′=sin x,
11.26.а) x2 y′′= y′2 ,
11.27.а) 2xy′y′′= y′2 −4,
11.28.а) y′′′xln x = y′′,
11.29.а) y′′c tgx + y′ = 2,
11.30.а) (1+ x2 ) y′′= 2xy′,
б) y′′+ 2 y′2 = 0. 1− y
б) y′′(1+ y) =5y′2 .
б) y′′(2 y +3) −2 y′2 = 0.
б) 4 y′′2 =1+ y′2 . б) 2 y′2 =( y −1) y′′. б) 1+ y′2 = yy′′. б) y′′+ yy′3 = 0. б) yy′′− y′2 = 0.
б) yy′′− y′2 = y2 ln y.
б) y (1−ln y) y′′+(1+ln y) y′2 = 0.
б) y′′(1+ y) = y′2 + y′. б) y′′= yy′ .
б) y′′=1+2y′2 .
б) yy′′−2 yy′ln y = y′2 .
б) y′′= 1y .
Задача 12. Найти решение задачи Коши.
12.1. y′′+3y′+ 2 y = 0, y(0) =1, |
y′(0) = −2. |
12.2. y′′− y′−2 y = 0, y(0) = −1, |
y′(0) =3. |
12.3. 5y′′+3y′= 0, y(0) = 4, y′(0) =1. |
|
12.4. y′′−5y′+6 y = 0, y(0) = 0, |
y′(0) =5. |
101
12.5. y′′+ y′−2 y = 0, y(0) = 2, |
y′(0) = −2. |
||
12.6. y′′− y′−12 y = 0, y(0) = 2, |
|
y′(0) = 0. |
|
12.7. y′′+ y′−6 y = 0, y(0) =3, |
y′(0) = −1. |
||
12.8. y′′−49 y = 0, y(0) = −5, y′(0) = 2. |
|||
12.9. y′′−5y′+ 4 y = 0, y(0) = 0, |
|
y′(0) = −4. |
|
12.10. y′′−6 y′+8y = 0, y(0) =5, |
y′(0) =1. |
||
12.11. 4 y′′−8y′+3y = 0, y (0) = 2, |
y′(0) = 0. |
||
12.12. y′′−3y′−10 y = 0, y(0) = −3, |
y′(0) = 2. |
||
12.13. y′′−4 y′−21y = 0, y(0) = 0, |
y′(0) = 4. |
||
12.14. 2 y′′+3y′+ y = 0, y(0) =1, |
y′(0) =1. |
||
12.15. y′′+ 4 y′= 0, y(0) = −3, y′(0) = −1. |
|||
12.16. y′′−10 y′+ 21y = 0, y(0) = 2, |
y′(0) = 0. |
||
12.17. y′′−25y = 0, y(0) = −1, y′(0) = 2. |
|||
12.18. y′′−7 y′−8y = 0, y(0) = 0, |
y′(0) = −3. |
||
12.19. y′′−3y′−4 y = 0, y(0) = 2, y′(0) = −4.
12.20. |
y′′−10 y′+16 y = 0, y(0) = 0, |
y′(0) = 7. |
12.21. |
y′′−3y′−18y = 0, y (0) = −2, |
y′(0) =1. |
12.22. |
y′′−2 y′−15y = 0, y(0) =3, |
y′(0) = 2. |
12.23. |
y′′−5y′= 0, y(0) =1, y′(0) = 4. |
|
12.24. |
y′′−7 y′+6 y = 0, y(0) = −2, |
y′(0) = −3. |
12.25. |
y′′+6 y′+8y = 0, y(0) =1, y′(0) = −2. |
12.26. |
y′′−16 y = 0, y (0) = −2, y′(0) = 2. |
12.27. |
4 y′′+8y′−5y = 0, |
y(0) =3, |
y′(0) = 0. |
12.28. |
9 y′′+3y′−2 y = 0, |
y(0) = 2, |
y′(0) =1. |
102
12.29. |
6 y′′+7 y′−3y = 0, y (0) = 0, |
y′(0) = 4. |
12.30. |
4y′′−8y′+3y = 0, y(0) = 2, |
y′(0) = 0. |
Задача 13. Найти общее решение уравнений.
13.1.а) y′′−10 y′+ 25y = 0,
13.2.а) y′′+ 4 y′+ 4 y = 0,
13.3.а) y′′−6 y′+9 y = 0,
13.4.а) 4 y′′−12 y′+9 y = 0,
13.5.а) y′′−8y′+16 y = 0,
13.6.а) 9 y′′−30 y′+ 25y = 0,
13.7.а) y′′+8y′+16 y = 0,
13.8.а) 25y′′−10 y′+ y = 0,
13.9.а) y′′+12 y′+36 y = 0,
13.10.а) y′′−2 y′+ y = 0,
13.11.а) 16 y′′+ 24 y′+9 y = 0,
13.12.а) y′′+14 y′+ 49 y = 0,
13.13.а) 9 y′′+6 y′+ y = 0,
13.14.а) y′′+6 y′+9 y = 0,
13.15.а) 49 y′′+70 y′+ 25y = 0,
13.16.а) y′′−4 y′+ 4 y = 0,
13.17.а) 4 y′′+12 y′+9 y = 0,
13.18.а) 9 y′′+30 y′+ 25y = 0,
13.19.а) 16 y′′−24 y′+9 y = 0,
13.20.а) y′′−10 y′+ 25y = 0,
13.21.а) 64 y′′+16 y′+ y = 0,
13.22.а) 25y′′−70 y′+ 49 y = 0,
13.23.а) 9 y′′−24 y′+16 y = 0,
б) y′′+ 4 y = 0.
б) y′′−4 y′+13y = 0. б) y′′+9 y = 0.
б) y′′+ 2 y′+5y = 0.
б) y′′+144 y = 0.
б) y′′+ 2 y′+17 y = 0. б) y′′−4 y′+ 20 y = 0. б) y′′−4 y′+5y = 0. б) y′′+ 25y = 0.
б) y′′+ 4 y′+13y = 0. б) y′′+ 2 y′+10 y = 0. б) y′′+ 4 y′+ 20 y = 0. б) y′′+ y = 0.
б) y′′+ 4 y′+8y = 0. б) y′′+64 y = 0.
б) y′′+10 y′+ 29 y = 0. б) y′′−6 y′+34 y = 0. б) y′′−4 y′+13y = 0. б) y′′+6 y′+13y = 0. б) y′′−4 y′+53y = 0. б) y′′+ 2 y′+5y = 0.
б) y′′+ 256 y = 0.
б) y′′+6 y′+ 25y = 0.
103
13.24. |
а) 4 y′′+ 4 y′+ y = 0, |
б) |
y′′+6 y′+34 y = 0. |
13.25. |
а) 9 y′′+12 y′+ 4 y = 0, |
б) |
y′′−2 y′+10 y = 0. |
13.26. |
а) 4 y′′+ 20 y′+ 25y = 0, |
б) |
y′′+144 y = 0. |
13.27. |
а) 25y′′+30 y′+9 y = 0, |
б) |
y′′−2 y′+17 y = 0. |
13.28. |
а) 36 y′′+12 y′+ y = 0, |
б) |
y′′−10 y′+ 29 y = 0. |
13.29. |
а) y′′−18y′+81y = 0, |
б) 4 y′′+9 y = 0. |
|
13.30. |
а) 4 y′′−44 y′+121y = 0, |
б) |
y′′−4 y′+8y = 0. |
Задача 14. Найти общее решение дифференциальных уравнений.
14.1.1) y′′′−3y′′=3x + x2
2)y′′′+3y′′−4 y′ =(x +1)ex
3)y′′−4 y′+ 4 y = e2 x sin 5x
14.2.1) y′′−6 y′+9 y = 2x2 + x +3
2)2 y′′′− y′′− y′= 4xe2 x
3)y′′+ 49 y = 2cos 7x −3sin 7x
14.3.1) 3y′′′− y′′ = 6x −1
2)y′′′+6 y′′+9 y′=(x −1)e−3x
3)y′′−4 y′+ 4 y = −e2 x sin 6x
14.4.1) y′′′−5y′′+6 y′=(x −1)2
2)y′′′−7 y′′=5xex
3)y′′+ 2 y′=10ex (sin x +cos x)
14.5.1) y′′′+ y′′−2 y′= −8x3
2)y′′′−3y′′+ 2 y′ =(1−2x)ex
3)y′′+ 2 y′+5y = −xsin 2x
104
14.6.1) y′′′−13y′′+12 y′ = x −1
2)y′′′+ y′′−6 y′=(20x +14)e2 x
3)y′′−4 y′+ 4 y = e2 x sin 3x
14.7.1) y′′′−2 y′′= −6( x +1)
2)y′′′−4y′′+3y′= −4xex
3)y′′+6 y′+13y = e−3x cos5x
14.8.1) y′′−6 y′+8y =3x2 + 2x +1
2)y′′′+ y′′−12 y′=(16x + 22)e4 x
3)y′′+ 4 y = 2cos 2x +3sin 2x
14.9.1) yIV −2y′′′+ 2 y′′= 48x2
2)y′′′+3y′′=(40x +58)e2 x
3)y′′−4 y′+ 4 y = −e2 x sin 4x
14.10.1) y′′′− y′′−4 y′+ 4 y = x2 +3
2)y′′′−5y′′+8y′−4 y = e2 x
3)y′′+ 2 y′ =3ex (sin x +cos x)
14.11.1) yIV + 2 y′′′+ y′′= 2 −3x2
2)y′′′−3y′′+ 2 y′=(x2 + x)e3x
3)y′′+ y = 2cos3x −3sin 3x
14.12.1) y′′+ 2y′+ 2 y = 2x2 +8x +8
2)y′′′+ y′′ =(1−4x)e−2 x
3)y′′+ y = 2cos x
14.131) y′′+8y′+16 y =16x2 −16x +66
2)y′′′−6 y′′+9 y′= 4xe3x
3)y′′+ 2 y′ = 6ex (sin x +cos x)
105
14.14.1) y′′′+ y′′= 49 −24x2
2)y′′′− y′′− y′+ y =(3x +7)e2 x
3)y′′+9 y = 6cos3x
14.15.1) yIV −3y′′′+3y′′− y′= x −3
2)y′′′+ 4 y′′+3y′= 4(1− x)e−x
3)y′′−6 y′+ 25y = 2sin x +3cos x
14.16.1) y′′′− y′= x2 + x
2)y′′′−3y′′− y′+3y =(4 −8x)ex
3)y′′+6 y′+10 y =80ex cos x
14.17.1) yIV − y′′′=5(x + 2)2
2)y′′′+ y′′−2 y′ =(6x +5)ex
3)y′′−4 y′+13y = e2 x cos3x
14.18.1) y′′+12 y′= 6x2 + 2x +1
2)y′′′+ 2 y′′+ y′ =(18x + 21)e2 x
3)y′′+9 y =12cos3x
14.19.1) y′′−2 y′+5y =5x2 +6x −12
2)y′′′−3y′−2 y = −4xex
3)y′′−4 y′+ 4 y = e2 x sin 6x
14.20.1) y′′′+3y′′+ 2 y′=3 + 2x + x2
2)y′′′+3y′′+ 2 y′=(1−2x)e−x
3)y′′−9 y′+18y = 26cos x −8sin x
14.21.1) yIV + 2 y′′′+ y′′=12x2 −6x
2)y′′′− y′′−2 y′=(6x −11)e−x
3)y′′−2 y′+37 y =36ex cos6x
106
14.22.1) y′′′+ y′′=5x2 −1
2)y′′′−2 y′′+ y′=(2x +5)e2 x
3)y′′+ 2y′+5y = −8e−x sin 2x
14.23.1) y′′−3y′+ 2 y = 2x3 −30
2)y′′′+3y′′−10 y′= xe2 x
3)y′′−3y′+ 2 y = −sin x −7 cos x
14.24.1) y′′′−13y′′+12 y′=18x2 −39
2)y′′′−2 y′′−3y′= xe3x
3)y′′+ 2 y′ = 4ex (sin x +cos x)
14.25.1) yIV + 2 y′′′+ y′′= 4x2
2)y′′′−4 y′′+5y′−2 y =(16 −12x)e−x
3)y′′+ 2 y′+5y = −17sin 2x
14.26.1) y′′′+3y′′+ 2 y′=3x2 + 2x
2)y′′′−3y′′+ 4 y =(18x −21)e−x
3)y′′+16 y = 2cos 4x +3sin 4x
14.27.1) y′′′− y′= x2 + x
2)y′′′−3y′+ 2 y =(4x +9)e−2 x
3)y′′−4 y′+8y = ex (5sin x −3cos x)
14.28.1) 5y′′+9 y′−2 y = x3 −2x
2)y′′′+ 4 y′′+5y′+ 2 y =(12x +16)ex
3)y′′+ 25y =5sin 5x + 4cos5x
14.29.1) y′′′− y′=3x2 −2x +1
2)y′′′−2 y′′+ y′=(2x +5)e2 x
3)y′′+ y = 2cos 7x −3sin 7x
107
14.30.1) yIV + 4 y′′′+ 4 y′′= x − x2
2)y′′′+ 4 y′′+ 4 y′= xe−2 x
3)y′′+ 2 y′+5y = −2sin x
Задача 15. По заданным корням характеристического уравнения записать вид общего решения линейного неоднородного дифференциального уравнения.
15.1. |
k1 |
= 0, k2 |
= 2, k3,4 |
= ±2i |
а) |
f (x) = |
||
б) f (x) = |
||||||||
|
|
|
|
|
|
|||
15.2. |
k1 |
=1, k2,3 |
= 2 ±3i |
а) |
f (x) = |
|||
б) f (x) = |
||||||||
|
|
|
|
|
|
|||
15.3. |
k1 |
= −2, k2 = 2, k3,4 =1 |
а) |
f (x) = |
||||
б) f ( x) = |
||||||||
|
|
|
|
|
|
|||
15.4. |
k1 |
=i, k2 = −i, k3,4 = ±3 |
а) |
f (x) = |
||||
б) f (x) = |
||||||||
|
|
|
|
|
|
|||
15.5. |
k1,2,3 = −1, k4,5 = 2 |
а) |
f (x) = |
|||||
б) f (x) = |
||||||||
|
|
|
|
|
|
|||
15.6. |
k1,2 |
= −2 ±i, k3,4 = 2 ±i |
а) |
f (x) = |
||||
б) f (x) = |
||||||||
|
|
|
|
|
|
|||
15.7. |
k1,2 |
= 0, k3,4 =1± |
4i |
а) |
f (x) = |
|||
б) f (x) = |
||||||||
|
|
|
|
|
|
|||
15.8. |
k1 |
= 0, k2 |
=1, k3,4 |
=3 ±i |
а) f (x) = |
|||
|
|
|
|
|
|
б) f (x) = |
||
15.9. |
k1 |
= −3, k2 = 2, k3 |
=5 |
а) |
f (x) = |
|||
б) f (x) = |
||||||||
|
|
|
|
|
|
|||
3sin 2x + 4 ; e−2 x (3x +5). 2sin 3x +е2 х ; x2ex .
e−2 x cos x + х ;
5xe2 x
2sin x −5cos x ; e−2 x (8x +3) −2.
e2 x (2sin 2x +cos 2x) ; 4xe−x −3.
e2x (cos x −3sin x) ; 2(1− x) +е−2 х.
3sin 4x +cos3x ; x2 −4x + 4.
e−3x sin x ; 5x2 +e2 x .
2sin 2x −3cos 2x; (x −5)2 −2е−3х.
108
15.10. |
k1 |
= 4i, k2 |
= −4i, k3,4 |
=5 |
а) |
f (x) = 2sin x +cos 4x ; |
|||
|
|
|
|
|
|
|
б) |
f (x) = x2 −25. |
|
15.11. k1,2,3 =3, k4,5 |
= ±5i |
|
а) f (x) = e3x (2x −3) ; |
||||||
|
б) |
f ( x) = e−x cos5x +3. |
|||||||
|
|
|
|
|
|
|
|||
15.12. |
k1,2 |
= −6 ±i, k3,4 = 0 |
|
а) |
f (x) = 2cos x +7sin 6x ; |
||||
|
б) f (x) = 6(1− x2 ). |
||||||||
|
|
|
|
|
|
|
|||
15.13. |
k1 |
= −3, k2 |
= 0, k3,4 = ±7i |
а) |
f (x) = −3cos7x; |
||||
|
|
|
|
|
|
|
б) f (x) = e−2 x (3x +5) +ех. |
||
15.14. |
k1 |
= 0, k2,3 |
= |
3 ±5i |
|
а) |
f (x) = 2e3x sin 5x ; |
||
|
б) f (x) = x3ex −е3х. |
||||||||
|
|
|
|
|
|
|
|||
15.15. |
k1 |
= 0, k2 = 2, k3,4 =1 |
а) |
f (x) = x +cos x ; |
|||||
б) f (x) =5xe2 x . |
|||||||||
|
|
|
|
|
|
|
|||
15.16. |
k1 |
=3i, k2 |
= −3i, k3,4 |
= ±1 |
а) |
f (x) = 2sin x −5cos x; |
|||
|
|
|
|
|
|
|
б) f (x) = ex (x +3) . |
||
15.17. |
k1,2,3 = −3, k4,5 = 0 |
|
а) f (x) = e−3x (2sin x +cos x); |
||||||
|
б) f (x) = 4x +1 |
||||||||
|
|
|
|
|
|
|
|||
15.18. k1,2 |
=1± 2i, k3,4 = −1± 2i |
а) |
f (x) = ex (cos 2x +sin 2x); |
||||||
б) f (x) =(2 − x)3 . |
|||||||||
|
|
|
|
|
|
|
|||
15.19. |
k1,2 |
= 0, k3,4 |
= ±4i |
|
а) |
f (x) =3sin 4x; |
|||
|
б) f (x) =(x2 −4x + 4)e4 x . |
||||||||
|
|
|
|
|
|
|
|||
15.20. k1 |
= −3, k2 |
= |
2, k3 = 0 |
а) |
f (x) = 2sin 2x +3cos x; |
||||
б) f (x) =(x −2)2 . |
|||||||||
|
|
|
|
|
|
|
|||
109
15.21. |
k1 |
= −2, k2 |
=1, k3,4 |
= |
3 ± 4i |
а) |
f (x) = e3x cos 4x; |
||
б) |
f (x) =5x . |
||||||||
|
|
|
|
|
|
|
|||
15.22. k1 |
= −4, k2 |
= 0, k3,4 |
= |
5 |
а) |
f (x) = 2sin x +e5x ; |
|||
б) f (x) = x2 −25 . |
|||||||||
|
|
|
|
|
|
|
|||
15.23. k1,2,3 = −1, k4,5 |
=3 |
|
|
а) f (x) = e3x (2x −3); |
|||||
|
|
б) |
f (x) = e−x cos5x . |
||||||
|
|
|
|
|
|
|
|||
15.24. |
k1,2 = −4 ±i, k3,4 = 0 |
|
а) |
f (x) = 2cos x +7sin x; |
|||||
|
б) f (x) = 6(1− x2 ) . |
||||||||
|
|
|
|
|
|
|
|||
15.25. |
k1 |
= 0, k2,3 = ±4i |
|
|
а) |
f ( x) =5cos 4x −3 ; |
|||
|
|
б) f (x) = e−4 x (2x +5). |
|||||||
|
|
|
|
|
|
|
|||
15.26. |
k1 |
= −3, k2,3 = |
2 ±5i |
|
а) |
f (x) = 2sin 5x +е2 х ; |
|||
|
б) f (x) = x2e−3x . |
||||||||
|
|
|
|
|
|
|
|||
15.27. |
k1,2 = ±5, k3,4 |
= 2 |
|
|
а) f (x) = e−2 x cos5x + 2 ; |
||||
|
|
б) f ( x) =8xe2 x . |
|||||||
|
|
|
|
|
|
|
|||
15.28. |
k1 |
= 2i, k2 |
= −2i, k3,4 |
= ±7 а) |
f (x) = 2sin 2x −5cos 2x ; |
||||
|
|
|
|
|
|
|
б) f (x) = 7e−2 x −2x −1. |
||
15.29. |
k1,2,3 = −3, k4,5 |
=3 |
|
|
а) |
f (x) = e3x (sin 3x −cos3x) ; |
|||
|
|
б) f (x) = 6xe−3x −3. |
|||||||
|
|
|
|
|
|
|
|||
15.30. |
k1,2 = −2 ± |
3i, k3,4 = 2 |
|
а) f (x) = e2x (cos x −3sin x) ; |
|||||
|
б) f (x) = 2(4 − x2 ) +9е2 х. |
||||||||
|
|
|
|
|
|
|
|||
110