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

The solution $${{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} + 17y = 0;$$ $$y\left( 0 \right) = 1,{\left( {{{d\,y} \over {d\,x}}} \right)_{x = {\raise0.5ex\hbox{$\scriptstyle \pi $} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}}} = 0\,\,$$ in the range $$0 < x < {\pi \over 4}$$ is given by