Variable separable

If $y = y(x)$ satisfies the differential equation

$$8\sqrt{x}\left(\sqrt{9 + \sqrt{x}}\right)dy = \left(\sqrt{4 + \sqrt{9 + \sqrt{x}}}\right)^{-1}dx, \quad x > 0$$

and $y(0) = \sqrt{7}$, then $y(256) =$

If $(2+\sin x)\dfrac{dy}{dx}+(y+1)\cos x=0$ and $y(0)=1$, then $y\left(\dfrac\pi2\right)$ is equal to :

A tangent to the curve, $y=f(x)$ at P(x, y) meets $x$-axis at A and $y$-axis at B. If AP : BP = 1 : 3 and $f(1)=1$, then the curve also passes through the point :

For x ∈ R, x ≠ 0, if y(x) is a differentiable function such that $x\displaystyle\int_1^xy(t)\,dt=(x+1)\int_1^xt\,y(t)\,dt$, then y(x) equals :

(where C is a constant.)

If $y(x)$ is the solution of the differential equation $(x+2)\dfrac{dy}{dx}=x^2+4x-9$, $x\ne-2$ and $y(0)=0$, then $y(-4)$ is equal to :

Let $f:[0,2]\to\mathbb R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$. Let
$$F(x)=\int_0^{x^2}f(\sqrt t)\,dt$$
for $x\in[0,2]$. If $F'(x)=f'(x)$ for all $x\in(0,2)$, then $F(2)$ equals

Let $f$ be a non-negative function defined on the interval $[0,1]$. If
$$\int_0^x\sqrt{1-(f'(t))^2}\,dt=\int_0^xf(t)\,dt,\qquad0\le x\le1,$$
and $f(0)=0$, then

Match the statements/expressions in Column I with the open intervals in Column II.

Let $f:\mathbf R\to\mathbf R$ be a continuous function which satisfies
$$f(x)=\int_0^xf(t)\,dt.$$
Then the value of $f(\ln5)$ is

Let a solution $y=y(x)$ of the differential equation
$$x\sqrt{x^2-1}\,dy-y\sqrt{y^2-1}\,dx=0$$
satisfy $y(2)=\frac2{\sqrt3}$.

STATEMENT-1: $y(x)=\sec\left(\sec^{-1}x-\frac\pi6\right)$

and

STATEMENT-2: $y(x)$ is given by $\dfrac1y=\dfrac{2\sqrt3}{x}-\sqrt{1-\dfrac1{x^2}}$

The differential equation $\dfrac{dy}{dx}=\dfrac{\sqrt{1-y^2}}{y}$ determines a family of circles with