Consider the differential equation $$\frac{d^2y}{dx^2}-2\tan x \frac{dy}{dx}-y=0$$ defined on $\big(-\frac{\pi}{2}, \frac{\pi}{2}\big)$. Which among the following are true?
there is exactly one solution $y=y(x)$ with $y(0) = y'(0) = 0$ and $y\big(\frac{\pi}{3}\big) = 2\big(1+\frac{\pi}{3}\big)$
there is exactly one solution $y=y(x)$ with $y(0) =1, \ y'(0) = -1$ and $y\big(-\frac{\pi}{3}\big) = 2\big(1+\frac{\pi}{3}\big)$
any solution $y=y(x)$ satisfies $y''(0) = y(0)$.
if $y_1$ and $y_2$ are any two solutions then $(ax+b)y_1 = (cx+d)y_2$ for some $a, b, c, d \in \mathbb{R}$.
How to solve the above problem?
Option 3 Just check ... $$y''(0)=y(0) \implies -2\tan (0)y'(0)=0 \text { and } \tan(0)=0$$ Do the same for option 4 Plug the solution given $y_1,y_2$ and try to find a,b,c,d...
Note that ( even you don't need the exact solution for this exercice )
$$\frac{d^2y}{dx^2}-2\tan x \frac{dy}{dx}-y=0$$ $$\cos(x)\frac{d^2y}{dx^2}-2\sin x \frac{dy}{dx}-y\cos(x)=0$$ $$\cos(x)\frac{d^2y}{dx^2}-\sin x \frac{dy}{dx}-\sin x \frac{dy}{dx}-y\cos(x)=0$$ $$(\cos(x)y')'-(\sin (x )y)'=0$$ You can integrate it easily $$(\cos(x)y')-(\sin (x )y)=K_1$$ $$(\cos(x)y)'=K_1 \implies \cos(x)y=K_1x+K_2$$