The order of the differential equation of all parabolas whose axis of symmetry is along the $X$-axis is:

Verify that the given function $x+y=\tan ^{-1} y$ is a solution of the differential equation $y^{2} y^{\prime}+y^{2}+1=0$.

The differential equation for which $y^2 = 4a(x+a)$ (where $a$ is the parameter) is the general solution is:

The differential equation satisfied by the family of curves $y = ax \cos \left( \frac{1}{x} + b \right)$,where $a$ and $b$ are parameters,is

Let $f(x)$ be a positive function such that the area bounded by $y=f(x), y=0$ from $x=0$ to $x=a>0$ is $e^{-a}+4a^2+a-1$. Then the differential equation,whose general solution is $y=c_1 f(x)+c_2$,where $c_1$ and $c_2$ are arbitrary constants,is :

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