The degree of the differential equation obtained from the equation $(y - a)^2 = 4(x - b)$ [where $a$ and $b$ are arbitrary constants] is

$y = e^x (A \cos x + B \sin x)$ is the solution of the differential equation

The differential equation of the family of curves $v = \frac{A}{r} + B$,where $A$ and $B$ are arbitrary constants,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 obtained by eliminating the arbitrary constants $a$ and $b$ from $xy = ae^x + be^{-x}$ is

By eliminating the arbitrary constants from $y=(a+b) \sin (x+c)-d e^{x+e+f}$,the differential equation obtained is of order