Differential Equations

A curve passes through the point $\left(1,\frac\pi6\right)$. Let the slope of the curve at each point $(x,y)$ be $\dfrac yx+\sec\left(\dfrac yx\right)$, $x>0$. Then the equation of the curve is

If $y(x)$ satisfies the differential equation $y'-y\tan x=2x\sec x$ and $y(0)=0$, then

Let $f:[1,\infty)\to[2,\infty)$ be a differentiable function such that $f(1)=2$. If
$$6\int_1^xf(t)\,dt=3xf(x)-x^3$$
for all $x\ge1$, then the value of $f(2)$ is

Let $y'(x)+y(x)g'(x)=g(x)g'(x)$, $y(0)=0$, $x\in\mathbb R$, where $f'(x)$ denotes $\dfrac{df(x)}{dx}$ and $g(x)$ is a given non-constant differentiable function on $\mathbb R$ with $g(0)=g(2)=0$. Then the value of $y(2)$ is

Let f be a real-valued differentiable function on $\mathbb R$ (the set of all real numbers) such that $f(1)=1$. If the y-intercept of the tangent at any point P(x, y) on the curve $y=f(x)$ is equal to the cube of the abscissa of P, then the value of $f(-3)$ is equal to

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.

Match the statements/expressions given in Column I with the values given 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 $\dfrac1y=\dfrac{2\sqrt3}{x}-\sqrt{1-\dfrac1{x^2}}$

Let $f(x)$ be differentiable on the interval $(0,\infty)$ such that $f(1)=1$, and
$$\lim_{t\to x}\frac{t^2f(x)-x^2f(t)}{t-x}=1$$
for each $x>0$. Then $f(x)$ is

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