The differential equation of all parabolas each of which has a latus rectum $4a$ and whose axes are parallel to the $x$-axis is:

Verify that the given function $y=\sqrt{1+x^{2}}$ is a solution of the differential equation $y^{\prime}=\frac{xy}{1+x^{2}}$.

Let the tangent to a curve at point $(x, y)$ intersect the $Y$-axis at point $P$. $A$ line drawn through point $P$ is perpendicular to this tangent and passes through the point $(1, 0)$. The differential equation of the curve is:

If the order of a differential equation $\frac{d^2 y}{d x^2}-2\left(\frac{d y}{d x}\right)^3+\sin \left(\frac{d y}{d x}\right)+y=0$ is $l$ and the degree of the differential equation $\left(1+\frac{d^2 y}{d x^2}\right)^{\frac{2}{3}}=\left[2-\left(\frac{d y}{d x}\right)^3\right]^{\frac{3}{2}}$ is $m$, then the differential equation corresponding to the family of curves $y=A x^l+B e^{m x}$, where $A$ and $B$ are arbitrary constants, is

The differential equation of a family of hyperbolas whose axes are parallel to coordinate axes, centres lie on the line $y=2x$ and eccentricity is $\sqrt{3}$ is

The differential equation for the line $y = mx + c$ is (where $m$ and $c$ are arbitrary constants).