Question

Solve the equation:
u′′+4u+5u=0

With the initital values:
u(0)=1
u′(0)=0

Answer 1

u′′ wht's this? O.o

Answer 2

d^2y/dx^2

Answer 3

Aux equation is:
\[\lambda^{2}+4 lambda + 5=0\]

Answer 4

\[\lambda = - 2 \pm i\]

Answer 5

Complex conjugate roots, keep going..

Answer 6

Woops made a mistake in the initital equation should be:
\[u''+4u'+5u=0\]

Answer 7

I guessed that...:-

Answer 8

So general solution is...?

Answer 9

Then the general equation is:
\[u = Ae ^{\lambda}+Be^{\lceil lamdba \rceil}\]
no complex conjugate symbol so I used ceiling

Answer 10

no

Answer 11

\[e^{-2t}(Acos(t)+Bsin(t))\]

Answer 12

thats general soln

Answer 13

then its just a ton of product rules with the conditions to find the constants

Answer 14

which can be written:
\[u=Ae^{\alpha x}e^{i \beta x} + Be^{\alpha x}e^{-i \beta x}\]

Answer 15

so finding it for u I just plug in zero

Answer 16

Yes, elecengineer has just substituted in already...

Answer 17

and make it equal to 0

Answer 18

Think you have gone off the track.

Go back to where elecengineer put up the general solution (in t instead of u).

Answer 19

assuming u as function of t

general soln

\[u(t) = e^{-2t}(Acos(t) +Bsin(t) )\]

Answer 20

u(0) = 1

gives 1= 1( A)

ie A=1

Answer 21

then differenitate

du/dt = e^(-2t) ( -sin(t) +Bcos(t) ) +(-2e^(-2t) ) ( cos(t) +Bsin(t) )

Answer 22

\[\frac{du}{dt} = e^{-2t} ( -Asin(t) +Bcos(t) ) -2e^{-2t}(Acos(t) +Bsin(t) ) \]

Answer 23

sub A=1, du/dt =0 , t=0

Answer 24

0 = 1(0+B) - 2(1)(1)

Answer 25

B = 2

Answer 26

B=2

soln is \[u(t) = e^{-2t} ( \cos(t) +2\sin(t))\]

Answer 27

Sweet thanks

Answer 28

Really helped with my revision