The equation of a curve whose normal at any point has a slope which is the same as the ordinate and which passes through $(1, -1)$ is $2x = k(3 - y^2)$. Then $k$ is:

The assets of a person are increasing at a rate proportional to the square root of the assets at a given time. His assets increase from Rupees $9$ crores to Rupees $16$ crores in $2$ years,then his assets at the end of $5$ more years will be

The area bounded by a curve,the $x$-axis,and the ordinate of some point $(x, y)$ of the curve is equal to the length of the corresponding arc of the curve from a fixed point. If the curve passes through the point $P(0, 1)$,then the equation of this curve is:

If the tangent at a point $P(x, y)$ of a curve is perpendicular to the line that joins the origin with the point $P$,then the curve is

Let $\int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} dt = \int_0^x y(t) dt, 0 \leq x \leq 3, y \geq 0$,$y(0)=0$. Then at $x=2, y^{\prime \prime}+y+1$ is equal to :

Let the curve $y = y(x)$ be the solution of the differential equation,$\frac{dy}{dx} = 2(x + 1)$. If the numerical value of the area bounded by the curve $y = y(x)$ and the $x-$axis is $\frac{4\sqrt{8}}{3}$,then the value of $y(1)$ is equal to