Question

Differential equations

Answer 1

give question......

Answer 2

wt of them??

Answer 3

\[y"+3y'-4y = 0\]

Answer 4

if we let \[y= e^{mx}\]

Answer 5

I know that we can get

\[(m^2 +3m-4)e^{mx}=0\]

Answer 6

use characteristic equation:
m^2+3m-4=0
m=-4 and m=1
so
y=c1e^(-4t)+c2e^t

Answer 7

(d^2+3d-4)y=0
d=4,-1
y=ae^4x+be^-x

Answer 8

and it factors into

\[(m-1)(m+4)=0\]

Answer 9

so since the roots are real and distinct,

y = \[ae^{x}+be^{-4x}\]

Answer 10

CORRECT

Answer 11

it will be ae^4x+be^-x.

Answer 12

but what if

\[y"-10y'+25y = 0 \]

Answer 13

because your char. equation m^2+3m-4=0 has the roots 4,-1.

Answer 14

(a+bx)e^5x

Answer 15

Then you have two same roots m=5
and then
y=c1te^5t+c2e^5t

Answer 16

how do we proceed the problem?

do we let

\[y= xe^{mx}\]

Answer 17

"because your char. equation m^2+3m-4=0 has the roots 4,-1" dipank this is NOT CORRECT

Answer 18

yes i checked........sorry for that........

Answer 19

noproblem, i just didn't want to confuse..

Answer 20

for the other root I think there was another way to solve it

when the roots are equal and real, the formula was different

and I think dipank sounds right

but I am not sure how to prove it

Answer 21

long proof..it'd be in your book for sure. circuit theory is the only way i ever wrapped my brain around this concept..a critically damped RLC circuit is modelled by a characteristic equation with two real same roots.
Good luck with your proof and further studies though.

Answer 22

here is a link that might help you understand repeated roots in characteristic equation

http://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx