Solve the differential equation y"+4y=0.
r^2+4=0
r=2i, -2i
Now what?
Solve the differential equation y"+4y=0.
r^2+4=0
r=2i, -2i
Now what?
When you get complex roots a ± bi, the general solution becomes y(t) = e^{at}cos(bt) + e^{at}sin(bt). So for this problem you get y(t) = cos(2t) + sin(2t)
You have applied Euler's method to solve the homogeneous with constant coefficeints.
The common solution of the equation can be expressed like this:
y = C_{1} e ^{rx} + C_{2} e ^{-rx } (1)
You have found the right equation for r after substituting (1) into you differential equation. Use
r = 2i and r = -2i
to come up with final solution:
y = C_{1} e ^{ 2ri } + C_{2} e ^{- 2ri}
This solution can be expressed in terms of trigonometric functions, but coefficients C_{1} and C_{2} have to be found from boundary conditions which are not given here.