The solution of the ordinary differential equation $${{dy} \over {dx}} + 2y = 0$$ for the boundary condition, $$y=5$$ at $$x=1$$ is

The solution to the ordinary differential equation $${{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} - 6y = 0\,\,\,$$ is

The solution for the differential equation $$\,{{d\,y} \over {d\,x}} = {x^2}\,y$$ with the condition that $$y=1$$ at $$x=0$$ is

Transformation to linear form by substituting $$v = {y^{1 - n}}$$ of the equation $${{dy} \over {dt}} + p\left( t \right)y = q\left( t \right){y^n},\,\,n > 0$$ will be