This document contains solutions to 14 differential equation problems. For each problem, the key steps taken are to find the roots of the differential equation, determine the appropriate solution case based on the roots, find values for constants by substituting initial conditions, and arrive at the solution. Most problems are solved by finding roots, determining the case is 1 or 3 based on real or imaginary roots, then solving for constants to match initial conditions.

1. 0
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Problems Collection
Differential Equation II
Meiva Marthaulina
4103312019

2. 1
Determine the solution of :
1.
With y(0) = 0 and y’(0) = 1
Solution :
Eq :
To find the root
To determine the case
Since so we take case 1 that
To determine y(0) = 0
... (1)
To determine y’(0)=1
y’(t)=
y’(0)=1
... (2)
To subsitute (1) to (2)

3. 2
Conclusion
So, we get :
2.
With y(0) = 0 and y’(0) = 1
Solution :
Eq :
To find the root
To determine the case
Since so we take case 1 that
To determine y(0) = 0
... (1)
To determine y’(0)=1
y’(t)=
y’(0)=1

4. 3
... (2)
To subsitute (2) to (1)
Conclusion
So, we get :
3.
With y(0) = 0 and y’(0) = 1
Solution :
Eq :
To find the root
To determine the case
Since so we take case 1 that
To determine y(0) = 0
... (1)

5. 4
To determine y’(0)=1
y’(t)=
y’(0)=1
... (2)
To subsitute (1) to (2)
Conclusion
So, we get :
4.
Solution :
Eq :
To find the root

6. 5
To determine the case
Since the root is imaginer so we take case 3 that
Conclusion
So, we get :
5.
Solution :
Eq :
To find the root
To determine the case
Since the imaginer root so we take case 3 that

7. 6
Conclusion
So, we get :
) +
6.
Solution :
Eq :
To find the root
To determine the case
Since is real number so we take case 1 that and is imaginer and
take case 3
Conclusion
So, we get :
7.
Solution :
Eq :

8. 7
To find the root
To determine the case
Since the imaginer root so we take case 3
Conclusion
So, we get :
8.
Solution :
Eq :
To find the root

9. 8
To determine the case
Since the imaginer root so we take case 3
Conclusion
So, we get :
9.
y(0) = 0 , y’(0) = 1 , y’’(0) = 2
solution :
Eq :
To find the root
To determine the case
To find the value of

10. 9
To find constanta
y(0) = 0
0 =
... (1)
y’(0) = 1
... (2)
y’’(0) = 2
... (3)
Doing elimination
(1) and (3)
-

11. 10
Equation 1
Equation 2
Conclusion
So, we get :
10.
y(0) = y’(0) = y’’(0) = y’’’(0) = 1
solution :
eq :
To find the root
By horner method, we get the roots are 3 and -3, so we have :

12. 11
To determine the case
To find the value of
To find constanta
y(0) = 0
0 =
... (1)
y’(0) = 1
... (2)
y’’(0) = 1

13. 12
... (3)
y’’’(0) = 1
... (4)
Doing elimination
(1) and (2)
-
... (5)
(5) and (3)
-
... (6)
(2) and (3)

14. 13
-
... (7)
(1) and (7)
-
... (8)
(8) and (6)
-
Subtitute to equation (6)
Subtitute to equation (1)

15. 14
Subtitute to equation (3)
Conclusion
So, we get :
11.
q(0) = q’(0) = 0
solution :
Eq :
To find the root

16. 15
To determine the case
To find the value of
To find constanta
y(0) = 0
0 =
... (1)
y’(0) = 0
... (2)

17. 16
Equation (1)
Conclusion
So, we get :
12.
y(0) = y’(0) = 0
solution :
Eq :
To find the root

18. 17
To determine the case
To find the value of
(1)
(2)
(3)
(4)

19. 18
To find constanta
y(0) = 1
1 =
... (1)
y’(0) = 1
... (2)
Equation (1)
Conclusion
So, we get :

20. 19
13.
solution :
Eq :
To find the root
To determine the case
To find the value of
(1)

21. 20
(2)
(3)
To find constanta
y(0) = 0
0 =
Conclusion
So, we get :

22. 21
14.
solution :
Eq :
To find the root
By horner method is gotten
, and
To determine the case
To find the value of
(1)

23. 22
(2)
(3)
To find constanta
y(0) = 0
0 =
... (1)
y’(0) = 0

24. 23
Equation (1) and (3)
__
Equation (1)
Equation (2)

25. 24
Conclusion
So, we get :