(d^2y/dx^2)+(9y)=0
y(0)=3 y'(0)=6
HOW DO YOU DO SECOND ORDER INITIAL VALUE
JUST TELL ME DONT DO IT

Question

(d^2y/dx^2)+(9y)=0
y(0)=3   y'(0)=6 
HOW DO YOU DO SECOND ORDER INITIAL VALUE
JUST TELL ME DONT DO IT

anonymous · Answer

is this calculus based?

anonymous · Answer

diffy Q

anonymous · Answer

dont know what that means,was just thinking it my be implicit derivates

anonymous · Answer

Differential Equations

anonymous · Answer

yeah! thats implicit differentiation then

amistre64 · Answer

y'' + 9y = 0

amistre64 · Answer

how to explain a characteristic equation ....

amistre64 · Answer

assume that e^(rx) will work

y = e^(rx)
y' = r e^(rx)
y'' = r^2 e^(rx); and sub it int

r^2 e^(rx) + 9e^(rx) = 0

factor out the es

e^(rx) (r^2 + 9) = 0


since e^rx is never zero; we solve for the quadratic

amistre64 · Answer

r^2 + 9 = 0

r^2 = -9

r = 0+3i

after that it gets funky :)  but it simplifies to:

$$y=e^{0x}(c_1cos(3x)+c_2sin(3x))$$
$$y=c_1cos(3x)+c_2sin(3x)$$