Differential Eq...

Solution of the equation $$\dfrac {dy}{dx} = \dfrac {y\dfrac {d(\phi (x))}{dx} - y^{2}}{\phi (x)}$$ is

The solution of the differential equation $$\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{\theta (y/x)}{\theta ' (y/x)}$$ is 

The solution of the differential equation
$$\left( 2x\cos { y } +3{ x }^{ 2 }y \right) dx+\left( { x }^{ 3 }-{ x }^{ 2 }\sin { y } -y \right) dy=0$$, is given by

The solution of differential equation $$x\cos^{2}y dx = y\cos^{2} x  dy$$ is

The solution of the differential equation
$$x=1+xy\cfrac { dy }{ dx } +\cfrac { { \left( xy \right)  }^{ 2 } }{ 2! } { \left( \cfrac { dy }{ dx }  \right)  }^{ 2 }+\cfrac { { \left( xy \right)  }^{ 3 } }{ 3! } { \left( \cfrac { dy }{ dx }  \right)  }^{ 3 }+...\quad \quad $$
is

Let f(x) be differentiable on the interval $$(0,\infty)$$ such that $$f(1)=1$$ and $$\displaystyle \lim_{t \rightarrow x }\dfrac{t^2f(x)-x^2f(t)}{t-x}=1$$ for each $$x>0$$. Then f(x) is 

The solution of the differential equation, $$x^2\dfrac{dy}{dx}.cos\dfrac{1}{x}=-1$$, where $$y\rightarrow -1$$ as $$x\rightarrow \infty$$ is 

Solution of the differential equation 
$$\frac{dy}{dx}$$ = $$\frac{3 x^2 y^4 + 2 xy}{x^2 - 2 x^3 y^3}$$ is

If x and y are independent variables and $$f(x)=(\int^x_0e^{x-y}f'(y)dy)-(x^2-x+1)e^x$$ where f(x) is a differentiable function, then $$f(\dfrac{-1}{2})$$ equals.

The solution of the differential equation $$\dfrac {x}{x^{2} + y^{2}} dy = \left (\dfrac {y}{x^{2} + y^{2}} + 1\right )dx$$ is