Latest Activity In Study Groups

Join Your Study Groups

VU Past Papers, MCQs and More

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

Views: 927

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.

sir also share this

Lecture 16 sa related hai

MTH401_GDB_Solution_Fall_2020

Click on the below link to download the file

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)

RSS

Looking For Something? Search Below

VIP Member Badge & Others

How to Get This Badge at Your Profile DP

------------------------------------

Management: Admins ::: Moderators

Other Awards Badges List Moderators Group

© 2021   Created by + M.Tariq Malik.   Powered by

Promote Us  |  Report an Issue  |  Privacy Policy  |  Terms of Service