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Q,$$\frac{d^2 \theta}{dt^2}=-k\theta $$ where $k=\frac{g}{l}$

How to solve this differential equation?

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$$\frac{d^2 \theta}{dt^2}+k\theta=0$$

According to this site, the auxiliary equation is

$$x^2+k=0$$ since,$$(0^{2}-4\times1\times\frac{g}{l})<0$$

Using from the site,

$$m_1 = α + jω$$

$$m_2 = α − jω$$

$$b^2 − 4ac<0$$ $${y}={e}^{{\alpha{x}}}{\left({A} \cos{\omega}{x}+{B} \sin{\omega}{x}\right)}$$

I can write that

$$\theta=e^{0\times\theta}(A cos{\sqrt{k} t})+B(sin \sqrt{k}t)$$

$$=(A cos{\sqrt{k}t})+B(sin \sqrt{k}t)$$

$$=A cos{\sqrt{\frac{g}{l}}t}+Bsin\sqrt{\frac{g}{l}}t$$

But wikipedia gives different solution on this page.

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3 Answers

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Using substitution $$ y=e^rx $$ $$ y'=re^rx $$ $$ y''=r^2 e^rx $$

and letting

$$\theta=y$$

then

$$y''=-ky$$ $$(r^2 e^rx)=-k(e^rx)$$ $$r^2=-k$$ $$r=\pm \sqrt{-k}$$ $$r=\pm i\sqrt{k}$$ $$r=0 \pm i\sqrt{k}$$ Which, when put into the characteristic equation for $r=\alpha+\beta i $, whose characteristic equation is $$y=e^{\alpha x} (C(1)\cos{\beta x} + C(2)\sin{\beta x}) $$

the general solution is $$y=C(1) \cos{\sqrt{k} x} + C(2) \sin{\sqrt{k} x} $$ where $C(1)$ and $C(1)$ are constants. Substituting back $y$ for $\theta$ and $x$ for $t$, and $k$ for $\frac{g}{l}$,

$$\theta=C(1) \cos{\sqrt{\frac{g}{l}} t} + C(2) \sin{\sqrt{\frac{g}{l}} t} $$

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To solve:

$$x''(t)=-\text{k}x(t)$$

Use Laplace transform:

$$\text{s}^2\text{X}(\text{s})-\text{s}x(0)-x'(0)=-\text{k}\cdot\text{X}(\text{s})$$

Solve $\text{X}(\text{s})$:

$$\text{X}(\text{s})=\frac{\text{s}x(0)+x'(0)}{\text{s}^2+\text{k}}$$

Using inverse Laplace transform:

$$x(t)=x(0)\cos\left(t\sqrt{\text{k}}\right)+\frac{x'(0)\sin\left(t\sqrt{\text{k}}\right)}{\sqrt{\text{k}}}$$

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Your solution is completely correct. I do not see where you see a difference to the Wikipedia page, the small angle solution is the same, only that the constants are inserted for the case starting at one of the extremal points where the velocity is zero.

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For large angles you have to use the equation for the physical pendulum, which the linked equation models. This does not have a solution that you can write down. As you can see on the wikipedia page, it is even difficult to compute the period of the oscillation.

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