www.vustudents.ning.com

We non-commercial site working hard since 2009 to facilitate learning Read More. We can't keep up without your support. Donate.

# MTH401 Differential Equations GDB Fall 2020 Solution / Discussion

MTH401 Differential Equations GDB Fall 2020 Solution / Discussion

Views: 928

### Replies to This Discussion

Share the GDB Question & Discuss Here....

Stay touched with this discussion, Solution idea will be uploaded as soon as possible in replies here before the due date.

Lecture 16 sa related hai

MTH401_GDB_Solution_Fall_2020

MTH401_GDB_Solution_Fall_2020

Right hai I have also same

GDB Solution:
Y=emx→yt=memx→yn=m2emx→ym=m3emx eq (i)
Let Y=emx→yt=memx→yn=m2emx→ym=m3emx
Put in eq (i) (m3emx)-2(m2emx) +4(memx)-8(emn) =0
emx(m3-2m2+4m-8) =0
m3-2m2+4m-8=0 eq (ii)
m3-2m2+4m-8=0 eq 1
p= -8=±1, ±2, ±4, ±8
q=1= ±1
p⁄q =±1, ±2, ±4, ±8
(2)3- 2 (2)2+4(2) - 8 =0
8-8+8-8=0
Put2 in eq 1 0=0
1 factor is (m-2)
And another factor find with synthetic division:
(m-2)=0
m2+4=0
m2=-4
m=0±2i
General solution is:
Y = c1emx+eax(c2cos 2x +c3sin 2x)
Y = c1e2x+ ( c2cos 2x +c3sin 2x)

1

2

3

4

5

## Latest Activity

2 hours ago
2 hours ago
+ M.Tariq Malik posted a discussion

### قرب کے نا وفا کے ہوتے ہیں - سارے جھگڑے انا کے ہوتے ہیں

2 hours ago
⭐ "Mannat"⭐✔️ posted a video

2 hours ago

2 hours ago

2 hours ago

2 hours ago

2 hours ago