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

If $${x^2}\left( {{{d\,y} \over {d\,x}}} \right) + 2xy = {{2\ln x} \over x}$$ and $$y(1)=0$$ then what is $$y(e)$$?

The solution of the differential equation $${{dy} \over {dx}} + {y^2} = 0$$ is

The equation $$\,\,\,{{{d^2}u} \over {d{x^2}}} + \left( {{x^2} + 4x} \right){{dy} \over {dx}} + y = {x^8} - 8\,\,{u \over {{x^2}}} = 8.\,\,\,$$ is a