Программная реализация метода прогонки и решение тестовых примеров. Структура системы линейных уравнений для сеточных значений функции

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Дифференциальное уравнение

Граничные условия

Точное решение

1

y''+y= 1

y'(0) = 0

y⎛⎜π⎞⎟− y'⎛⎜π⎞⎟= 2

⎝ 2 ⎠         ⎝ 2 ⎠

1+cosx

2

y''+2y'+2y= 2e^−x cosx

y'(0) = 0 y(π)=−e^−π

e^−x(cos x+ sin x+xsin x) 

3

y''−5y'+6y=e^x

y(0) = 0 b= ln5;  y'(b) =140

ex +e3x −e2x

4

y''+y=e^x +1

y(0) +y'(0) = 4 y(π) = 0.5e^π

cos x+ sin x+1+ 0.5e^x

5

y''+ ¹ y'− ¹₂ y= 8x x        x

y(1) + 2y'(1) =12

y(4) =

x³ + 2x+ ¹ x

6

x² y''+2.5xy'−y= 0

y(1) + 2y'(1) = 9

y(4) + y'(4) =

3 x − ¹₂

x

7

4xy''+2y'+y= 0

y(0) =1

y'(π² ) =− 2

sin x + cos x

8

x² y''−4xy'+6y= 2

y(1) − 2y'(1)  

y(5) − y'(5)

0.1x³ +x² +1/3

9

y''+4y= cos3x

5y(0) −y'(0) = 2

5y(π)−3y'(π) = 0

cos2x+sin 2x−

cos3x

10

x² y''−2y=10

y(1)+ y'(1) =−

− 8y(4) +16y'(4) = 39

1  ₂ 1 x + −5

2  3x

11

y''−2y'+y= xe^x

y(0)+ y'(0) = 3 y(1) − y'(1) =− exp(1)

e^x⎛⎜⎜x6³ +x+1^⎞⎟⎟⎠ ⎝

12

y''−2y'+y= 5xe^x

y(0)+ y'(0) = 3

  y(1) −y'(1) 1)

e^x +xe^x +x³e^x

13

y''+4y= sin x+sin2x

y(0)+ y'(

y'(π) =−

cos2x− sin 2x+

+sin x−xcos2x

14

y''+2y'+5y=17cos2x

−2y(0)+ y'(0) = 8 y(π) = 1

cos2x+ 4sin2x+e^−x sin2x

15

y''−2y'+y=e^x(sin x+1)

− 2y(0) +y'(0) =1 y(π) − y'(π) = −e^π(π+ 3)

ex⎛⎜ 1 x2 −sin x+ 2x⎞⎟

       ⎝ 2                        ⎠

16

x²y''+2xy'−6y = 24

y(1) +y'(1) = 1

y(4) − 4y'(4) =−

    2         ₂

+ 3x − 4

x3

17

x²y''−2xy'+2y =14

y(1) = 4 y(4) − 2y'(4) =−3

2x² −5x+ 7

18

2x² y''−3xy'+3y = 3

y(0) + y'(0) = 4 y(4) −y'(4) = 0

3x− 2x^3/2 +1

19

y''−2y'+2y = sin x

y(0) + 3y'(0) =1

y⎛⎜π⎞⎟= 1

⎝ 2 ⎠ 5

(sin x+ 2cos x)

20

y''+9y = 9

y(0) =−3 y(2π) +y'(2π) = 0

− 4cos3x+sin3x+1

21

y''+2y'+2y= 6

y(0) = 4 y(π) −y'(π) = 3

e^−x(cos x+ 2sin x)+ 3

22

y''−2y'+y =e^x

y(0) −y'(0) =−1

3y(3) −y'(3) =11exp(3)

e_x⎛⎜⎜x2² + x⎞⎟⎟⎠ ⎝

23

y''+ ² y'+y = 0 x

y'(π) =−

y(4π) =−

1

(2sin x−3cos x)

x

24

y''+4y'+5y=16cos x

y(0) = 2

y⎛⎜π⎞⎟+ 2y'⎛⎜π⎞⎟=−2

      ⎝ 2 ⎠         ⎝ 2 ⎠

2(cos x+sin x)

25

x²y''−xy'−3y= 3

y(0.5) + y'(0.5) =−22.75

y'(2.5) =−39.1

−2x³ +¹⁰−1 x