Differential Eq...

Solve: $$\dfrac{{dy}}{{dx}} = 1 + x + y + xy$$

The function f(x) satisfying the equation, $$f^2(x)+4f(x)\cdot f(x)+[f(x)]^2=0$$.where CCis an arbitrary constant.

$$\frac{{dy}}{{dx}} + \frac{{{y^2} + y + 1}}{{{x^2} + x + 1}} = 0$$

If $$y={e}^{4x}+2{e}^{-x}$$ satisfies the relation $$\dfrac {{d}^{3}y}{{dx}^{3}}+A\dfrac {dy}{dx}+By=0$$,then value of $$A$$ and $$B$$ respectively are 

The solution of the differential equation $$(1+y^{2})+(x-e^{\tan -1}y)\dfrac {dy}{dx}=0$$, is-

The solution of $$\dfrac {dy}{dx}+\dfrac {y}{x}=x^{2}y^{6}$$ is

The solution of the differential equation $$x^2\dfrac{dy}{dx}\cos\left(\dfrac{1}{x}\right) - y \sin\left(\dfrac{1}{x}\right) = -1 ; where \  y \rightarrow -1 \ as \ x \rightarrow \infty$$ is 

The general solution of differential equation $$\dfrac{dy}{dx}=\sin^{3}{x}\cos^{2}{x}+xe^{x}$$

The value of $$\displaystyle \lim_{x \rightarrow \infty}$$ y(x) obtained from the differential equation $$\dfrac{dy}{dx} = y - y^2$$, where y(0) = 2 is 

Solution of $$\left(\dfrac{x+y-a}{x+y-b}\right)\left(\dfrac{dy}{dx}\right)=\left(\dfrac{x+y+a}{x+y+b}\right)$$.