If $y=y(x)$ satisfies $\left(\frac{2+\sin x}{1+y}\right) \frac{dy}{dx} = -\cos x$ such that $y(0)=2$,then $y\left(\frac{\pi}{2}\right)$ is equal to

If $f(x)$ is a function such that $f^{\prime}(x)=\sqrt{f^2(x)-1}$ and $f(0)=1$,then $f(1)=$

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1$,with the initial condition $y(0)=1$. Then,the value of $\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ is equal to:

For solving $\frac{dy}{dx} = (4x + y + 1)$,the suitable substitution is:

The curve which passes through the point $(\sqrt{2}, 1)$ and satisfies the differential equation $\frac{dy}{dx} = \frac{2x}{3y}$ represents:

Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1, 2)$ and $(8, 1)$,then $\left| y \left(\frac{1}{8}\right) \right|$ is equal to