Question

find the general solution..
1. y''-2y'-2y=0

2. y''''-8y''-9y=0

3. y''''+13y''+36y=0

4. y'''''-15y''''+84y'''-220y''-275y'=0

5. y'''''+12y''''+104y'''+408y''-1156y'=0

Answer 1

Have you tried solving that one equation I asked you to solve?

Answer 2

no because its hard

Answer 3

y''+2y'+2y=0
r^2 + 2r + 2 = 0
r^2 + 2r = - 2
r^2 + 2r + 1 = - 2 + 1
(r + 1)^2 = - 1
r + 1 = ± √(-1)
r = - 1 ± i

the general solution is:

y = e^(-x)[c₁sin(x) + c₂cos(x)]
y = c₁e^(-x)sin(x) + c₂e^(-x)cos(x)

Answer 4

r^2-2r-2=0

Answer 5

Just use the quadratic formula.

Answer 6

from 2 to 5 cud u answer it?

Answer 7

wat # that i will use quadratc fmla?

Answer 8

Whenever you have something that can be written in the form ax^2+bx+c=0, you may use the quadratic formula.
You also may find synthetic division useful.

Answer 9

can u help me to solve this equation, if its ok 4 u....

Answer 10

Which one are you stuck on?

Answer 11

3,4,5...

Answer 12

So looking at number 3, we need to solve
\[y^{(4)}+13y^{(2)}+36y^{(0)}=0\]
means we need to look at how to solve
\[r^4+13r^2+36=0\]

Let u=r^2
so we have
\[u^2+13u+36=0\]
Use the quadratic formula or factor.

Answer 13

tnx how about #4 and # 5..just giv me a little bit of sol...and i'll be the one to continue to solv

Answer 14

Let me see your attempt on 4.

Answer 15

You should find one zero pretty easily.

Answer 16

can u answer it because i nid to sleep and this is my retrice. it will submit 2mrror mrning...f it is k 4 u...tnx in advance.....

Answer 17

myininaya?..pls

Answer 18

tnx 4 everything uve shared to me....ur a genius as wat i obsvd