Question

also please destribe which method you would use for solving this ODE: y"+9y=0; y(pie/3)=1, y'(pie/3)=-2

Answer 1

does seperation of church and state work on doubles?

Answer 2

just solve r^2+9=0 to figure out your solution form

Answer 3

thats what we call an imaginary solution

Answer 4

...and by we i mean me and my voices

Answer 5

mr.paul has the best site i have ever seen for talking about differential equations:
http://tutorial.math.lamar.edu/Classes/DE/HOHomogeneousDE.aspx

Answer 6

dy/dx + 9y = 0

dy/dx = -9y

dy/y = 9dx

ln(y) = 9x

Answer 7

\[y=c_1\cos(3x)+c_2\sin(3x)\]

Answer 8

i keep sleeping with my head on an ODE book, but i dont feel none to smarter

Answer 9

since \[r=\pm 3i \]

Answer 10

this isnt an exact equation?

Answer 11

supose a solution of the form:
\[y=e ^{ax}\]
replace in the equation:
\[a ^{2}e ^{ax}+ 9e ^{ax} =0\]
thus:
\[a ^{2}=-9\]
\[a=\pm3i\]
This entails:
\[y=Ae ^{3ix}+Be ^{-3ix}\]

Answer 12

if you have something in the form \[a_{n}y^{(n)}+a_{n-1}y^{(n-1)}+ \cdot \cdot \cdot +a_3y^{(3)}+a_2y''+a_1y'+a_0y=0\] we can find what our solution should be if we solve the polynomial:
(well as along as the a's are constant coefficients)
\[a_nr^n+a_{n-1}r^{n-1}+\cdot \cdot \cdot a_3r^3+a_2r^2+a_1r+a_0=0\]

Answer 13

for instance we had
\[y''+9y=0\]
this means all we really have to do
is solve
\[r^2+9=0=>r=\pm 3i \]
=>
\[y=Acos(3x)+Bsin(3x)\]
:)

if we have \[y''-9y=0\]
then solution is
\[y=ae^{3x}+be^{-3x}\]

since the solutions would have been r=pm 3

Answer 14

ooo gotcha

Answer 15

thank u so much

Answer 16

Refer to the attachment.