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MTH401 Differential Equations GDB Fall 2020 Solution / Discussion

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sir also share this

Lecture 16 sa related hai

MTH401_GDB_Solution_Fall_2020

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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)

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