$y=A e^x+B e^{2 x}+C e^{3 x}$ satisfies the differential equation

The differential equation of all circles which pass through the origin and whose centres lie on the $y$-axis is:

The differential equation of the family of curves $y=e^{x}(A \cos x+B \sin x)$,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

For each of the exercises given below,verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
$y=e^{x}(a \cos x+b \sin x) \quad: \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$

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