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In some physics problem I got next differential equation: x''=kx I know that simple harmonic oscillator problem are solved using complex numbers, but i don't know why solution is like: $x=A\cos\left(bt+d\right)$

I tried to solve this way.

$x''-kx=0$

$x=ae^{rt}$

$ar^2e^{rt}-kae^{rt}=0$

$r^2-k=0$

$$x=ae^{\pm\sqrt{k}t}$$

What did I do wrong?

Bonce
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1 Answers1

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The equation should be $x^{"}+kx=0$ where $k \ge 0$, in your case k should be negative and $r$ will have imaginary solutions so it brings sin and cos.

voyager
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