Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}=\frac{(\tan x)+y}{\sin x(\sec x-\sin x \tan x)}, \mathrm{x} \in\left(0, \frac{\pi}{2}\right)$ satisfying the condition $\mathrm{y}\left(\frac{\pi}{4}\right)=2$....

The solution curve of the differential equation $y \frac{d x}{d y}=x\left(\log _e x-\log _e y+1\right), x>0, y>0$ passing through the point $(e, 1)$ is

Let y=y(x) be the solution of the differential equation sec⁡xdy+{2(1-x)tan⁡x+x(2-x)} dx=0 such that y(0)=2. Then y(2) is equal to :

A function y=f(x) satisfies $f(x) \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0$ with condition $f(0)=0$. Then $f\left(\frac{\pi}{2}\right)$ is equal to

If the solution of the differential equation $(2 x+3 y-2) d x+(4 x+6 y-7) d y=0, y(0)=3$, is $\alpha x+\beta y+3 \log _e|2 x+3 y-\gamma|=6$,...

Let $x=x(t)$ and $y=y(t)$ be solutions of the differential equations $\frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0$ and $\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{by}=0$ respectively, $\mathrm{a}, \mathrm{b} \in \mathrm{R}$. Given that $x(0)=2 ; y(0)=1$ and $...

If for the solution curve $y=f(x)$ of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+(\tan x) y=\frac{2+\sec x}{(1+2 \sec x)^{2}}$, $x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$, then $f\left(\frac{\pi}{4}\right)$ is...

If $y=y(x)$ is the solution of the differential equation, $\sqrt{4-x^{2}} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\left(\left(\sin ^{-1}\left(\frac{x}{2}\right)\right)^{2}-y\right) \sin ^{-1}\left(\frac{x}{2}\right),-2 \leq x \leq 2, y(2)=\frac{\pi^{2}-8}{4}$, then $y^{2}(0)$ is equal...

Let $y = y(x)$ be the solution curve of the differential equation $...

Let $f:[0,\infty)\to$ $\mathbb{R}$ be a differentiable function such that $...