2nd order ODE $y''+2(\tan t)y'-y=0$ 
Let $I=(-\pi/2, \pi/2)$
Solve $y''+2(\tan t)y'-y=0$ with the initial conditions  $y(0)=2, y'(0)=1$

So far I've tried two methods.
The first method is what we saw in class: if you have an evident solution $\varphi$, then we can find the solution $\phi(t)=z(t)\varphi(t)$, where $z$ is a function such that $z'\varphi^2 = e^{-\int_{t_0}^t a(s)ds} $. In this case $a(s)=2\tan(s)$. But what could $\varphi$ be? I can't seem to see any "evident" solutions. I've tried $\sin t$ and $\cos t$ and I'm out of ideas because nothing seems to work.
The second method I've tried is transforming this into a first order ODE system. So we would have
$$x'=-2(\tan t )x+y$$
$$y'=x$$
But every time I try solving it using eigenvalues, the solution I get doesn't work.
 A: Hint.
Making $y(t) = (\sin t) u(t)$ we have
$$
 \sin t\cos t u'' + 2 u'=0
$$
which is separable after making $v = u'$ or
$$
\frac {dv}{v}+\frac{2dt}{\sin t\cos t}=0
$$
giving $\ln v + 2(\ln \sin t-\ln\cos t) = c_0$ or
$$
v\tan^2 t=c_1\Rightarrow u'=c_1\cot^2 t\Rightarrow u = c_2(t+\cot t)+c_3
$$
and finally
$$
y = (c_2(t+\cot t)+c_3)\sin t
$$
A: Consider that
$$
\left(p(t)y(t)\right)'' = p(t)y''(t)+2p'(t)y'(t)+p''(t)y(t)
$$
and compare with your equation to find a suitable factor $p(t)=\frac1{\cos(t)}$.

Taking the hint that the domain is $(-\pi/2,\pi/2)$, one could also try $y(t)=u(\sin(t))$, so that
$$
y'(t)=\cos t\,u'(\sin t),~~ y''(t)=\cos^2t\,u''(\sin t)-\sin t\,u'(t),
\\~\\
0=[\cos^2t\,u''(\sin t)-\sin t\,u'(t)]+2\sin t\,u'(t)-u(\sin t)
\\~\\
0=(1-x^2)u''(x)+xu'(x)-u
$$
As $u(x)=x$ is an obvious solution, setting $u(x)=xv(x)$ gives
$$
0=(1-x^2)[xv''(x)+2v'(x)]+x^2v'(x)=x(1-x^2)v''(x)+(2-x^2)v'(x)
$$
etc.
A: $$y''+2(\tan t)y'-y=0$$
Rewrite the DE as:
$$(1-u^2)y''+uy'-y=0$$
Where $u =\sin t$.
$y=u$ is a solution. You can reduce the order of the DE. The solution is not really nice.
