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?

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^{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.

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