The general solution of the differential equation $\left(x y+y^2\right) d x-\left(x^2-2 x y\right) d y=0$ is

The sum of the order and degree of differential equation $x\left(\frac{d^2 y}{d x^2}\right)^{1 / 2}=\left(1+\frac{d y}{d x}\right)^{4 / 3}$

$\frac{d y}{d x}=\frac{y+x \tan \frac{y}{x}}{x} \Rightarrow \sin \frac{y}{x}=$

$$ \begin{aligned} &\text { The general solution of the differential equation }\\ &(x+y) y d x+(y-x) x d y=0 \text { is } \end{aligned} $$

The general solution of the differential equation $$\frac{d y}{d x}=\cos ^2(3 x+y)$$ is $$\tan ^{-1}\left(\frac{\sqrt{3}}{2} \tan (3 x+y)\right)=f(x)$$. Then, $$f(x)=$$

If the general solution of the differential equation $$\cos ^2 x \frac{d y}{d x}+y=\tan x$$ is $$y=\tan x-1+C e^{-\tan x}$$ satisfies $$y\left(\frac{\pi}{4}\right)=1$$, then $$C=$$

Assertion (A) Order of the differential equations of a family of circles with constant radius is two.

Reason (R) An algebraic equation having two arbitrary constants is general solution of a second order differential equation.

If $$l$$ and $$m$$ are order and degree of a differential equation of all the straight lines at constant distance of $$P$$ units from the origin, then $$l m^2+l^2 m=$$

If $$2 x-y+C \log (|x-2 y-4|)=k$$ is the general solution of $$\frac{d y}{d x}=\frac{2 x-4 y-5}{x-2 y+2}$$, then $$C=$$

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

If the solution of $$\frac{d y}{d x}-y \log _e 0.5=0, y(0)=1$$, and $$y(x) \rightarrow k$$, as $$x \rightarrow \infty$$, then $$k=$$

$$y=A e^x+B e^{-2 x}$$ satisfies which of the following differential equations?

The solution of the differential equation $$2x\left(\frac{dy}{dx}\right)-y=4$$ represents a family of

The solution of the differential equation $$\frac{d^2 y}{d x^2}+y=0$$ is