Separable first order ODE involving tangent

[psie]

Homework Statement

Consider the DE $\frac{\mathrm{d}x}{\mathrm{d}t}=t\tan x$ with initial value $x(0)=\frac\pi{6}$.
1. Find the solution.
2. Describe the region in which the solutions are defined.

Relevant Equations

A separable first order ODE is of the form $x'=g(x)h(t)$.

By inspection, we see that $x=k\pi$ is a solution for $k\in\mathbb Z$. Moreover, the equation implicitly assumes $x\neq n\pi/2$ for odd $n\in\mathbb Z$, since $\tan x$ isn't defined there. So suppose $x\neq k\pi$, i.e. $\tan x\neq 0$, then rearranging and writing $\tan x=\frac{\sin x}{\cos x}$ we have $$\frac{x'\cos x}{\sin x}=\frac{\mathrm{d}}{\mathrm{d}t}\log(|\sin x|)=t.$$ Integrating and exponentiating, we obtain, $$|\sin x|=Ae^{\frac{t^2}{2}}$$ The initial condition implies $A=1/2$. Now, how do I get rid of the absolute values here? Does it somehow follow from the initial condition that $\sin x$ has to be positive?

 

[anuttarasammyak]

$$\sin x = \pm A e^{t^2/2}$$
where A >0. There are two cases of x>0 and x<0 during the time after. The initial condition says $x(0)=\pi/6 >0$ Our solution is the former one.
$$\sin x = \frac{e^{t^2/2}}{2}$$

 

[pasmith]

I think from $(\log |\sin x|)' = t$ the next step is $$\log |\sin x| = \log |A| + \tfrac12 t^2$$ and hence $$|\sin x| = |A|e^{\frac12 t^2}.$$ It follows from this that $\sin x$ and $A$ have the same sign, so we can drop the absolute value signs.