Задание 5 Найти решение краевой задачи

1. у – 5у = 30х – 11, у(0)=0, у(1) = 0

2. y+ 4y+4y= 6e^–2x,у(0) =1,у(1) =е^–2(y=e^–2x(1 – 3x+ 3x²)

3. y+ y = 2cos2x , y(0) = 1, y ( ) =1 (y = 0,6cosx + 0,4cos2x )

4. y – y = 2 –2x , y(0) =3, y (1) = e+2 ( y = e^x + x² +2 )

5. у –4у+ 4у = ,у(1) = ,у(0) =

6. y –8y + 20y= 130sin4x , y() = –2, y (0) = 0

7. y+4y –5y = 1,у(1)= 0,8,у(0)=0

8. y+ 6y +9y = ,у(1) = 0, у(0) = 0

9. y– 4y +5y = 2cosx , y(0)=,y () = –(y=e^2x(cosx–2sinx)+0,25(cosx–sinx))

10. y + y= 4х, y(0) = 1, y(1) = е ^–1

11. у – 4у + 4у = 3е^2х, у(2)= 0, у(0)=1

12. y +y = cosx + 2sinx, y (0)=0, y()=2 (y = –cosx +sinx +0,5xsinx –x cosx )

13. у – 5у = 7, y(1) = – 1,4 , y(0) = 0,6

14. у + 4у + 4у = 5е^–2х, у(2)= 5е ^–4, y (0) = 0

15. y+y =10e^xcosx, y( ) = 0, y(0)=7 ( y = cosx –2e^sinx +e^x(2cosx +4sinx))

Задание 6 Найти решение задачи Коши.

1. y+2y+y = 3cosx + 8sinx, у(0)=0, у(0)=

2. y + 4y = 8x³ + 1 ( y = C₁cos2x + C₂ sin2x + 2x³ –3x +0,25 )

3. y – y = 9xe^2x, y(0) = 0, y(0) = –5

4. y –4y+4y=sin2x,y(0)=,y(0) = 0

5. y + y = x² +1, y(0) = 1, y (0) =1 ( y = 2cosx + sinx +x² – 1 )

6. y– 5y+4y = (10х+43)е^–х, у(0) = у(0)=0,

7. у + 2у + у =3cosx +8sinx,

8. у + 4у = 3х – 2, y(0) = ,y (0) = 0,5

9. y – y = e^x , y(0) = 1, y (0) = 1,5 (y = e^x + 0,5xe^x )

10. у – 2у + у = sinx, y(0) = 0,5, y (0) = 2

11. y– 2y+5y = 5х, y(0)=0,6, y(0)= 0 (y = e^x(cos2x –sin2x) + х – 0,4)

12. y–3y +2y =xe^x , y(0) = 1, y (0) = –1 ( y = –e^2x +(2–0,5x² –x)e^x )

13. у – 6у + 9у = sin3x, y(0) = y(0)= 0

14. у+ 4у= 3х², y(0) =,y(0)= 2

15. y – 3y +2y = e^x,y(0) =2, y(0)= 0

16. у – 10у + 25у = 13cos x , y(0) = y(0)= 0

Задание 7 Решить дифференциальное уравнение методом Лагранжа.

1. y–2y+y=

2. y – 5y + 6y = e^x(e^x + 4) ( y = e^2x(x – 4e^–x +C₁)+e^3x(C₂ – e^–x –2e^-2x) )

3. y–2y = e^2x(e^x – 1)

4. y +2y +y = (y = (C₁ – x+xln|x|+C₂x)e^–x )

5. y+y=2ctg x

6. y –2y +y =(y = (x +xln|x| +C₂x+C₁)e^x )

7. y + ²y =

8. y + 4y+ 4y = (y =e^–2x(x –xlnx + ln|lnx| +C₁ +C₂ )

9. y – 9y+ 18y=

10. y – y =

11. y +3y =

12. y + 3y + 2y=

13. y + y + ctg²x= 0.

14. y – y =.

15. y + 5y + 6y=

16. y + 4y=

Задание 8

Решить дифференциальное уравнение, используя теорему наложения

1. y– 2y+10y=sin3x+e^x

2. y –2y –8y = e^x –8cos2x

3. y –6y +9y = .

4. y –2y +4y = .

5. y – 3y = x + cosx , y(0) =0, y =

6. y –2y +y = 2х + 3 + е^2х.

7. y +y = х+2е^х

8. y +3y + 2y = sin2x +2cos2x + 4

9. y +2y +y = 2х² + 3 + е^х.

10. y +2y +2y = + 4х.

11. y +3y= –72sin3x +1

12. y –4y +4y = е^3х +х² + х.

13. 5y –6y +5y = 5е^0,6х +

14. y –3y + 2y = 5sin2x +3х.

15. y –10y + 25y =

16. y + y = хе^х +2е^–х