If the solution $y=y(x)$ of the differential equation $\left(x^4+2 x^3+3 x^2+2 x+2\right) d y-\left(2 x^2+2 x+3\right) d x=0$ satisfies $y(-1)=-\frac{\pi}{4}$, then $y(0)$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $\left(1-x^2\right) d y=\left[x y+\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}\right] d x,-1< x<1, y(0)=0$. If $y\left(\frac{1}{2}\right)=\frac{m}{n}, m$ and $n$ are coprime...

If $x=x(t)$ is the solution of the differential equation $(t+1) d x=\left(2 x+(t+1)^4\right) d t, x(0)=2$, then, $x(1)$ equals

If the solution curve y=y(x) of the differential equation $\left(1+y^2\right)\left(1+\log _e x\right) d x+x d y=0, x>0$ passes through the point (1,1) and $...

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 $\alpha x+\beta y+3 \log _e|2 x+3 y-\gamma|=6$,...