# Thread: Simple second order d.e.

1. ## Simple second order d.e.

Sorry about this. I've solved this equation and I'm not quite getting what I am supposed to.

x''=-x can someone show me how to get the solution to this . Thanks

2. write as

x '' + x = 0

Assume solution of the form x = e^(rt)

Then plugging back into the original we obtain:

r^2e^(rt) + e^(rt) = 0

r^2 +1 = 0

This is the characteristic equation the solution is r = + i

The solutions are of the form e^(it) (you only need one of the complex solutions)

Using Euler's idenity e^(it) = cos(t) + i sin(t)

Then the real part: cos(t) and the imaginary part : sin(t) are the solutions

The general solution is x = Acos(t) + Bsin(t) which is easilyu verified to solve x '' + x = 0

3. Originally Posted by Mikey1205
Sorry about this. I've solved this equation and I'm not quite getting what I am supposed to.

x''=-x can someone show me how to get the solution to this . Thanks
The characteristic equation is $\displaystyle r^2=-1\implies r=\pm i$

Thus, the solution has the form $\displaystyle y=c_1\cos x+c_2\sin x$.