The solution of differential equation $${{dy} \over {dx}} = ky,y\left( 0 \right) = C$$ is

The order of differential equation $$\,\,{{{d^2}y} \over {d{t^2}}} + {\left( {{{dy} \over {dx}}} \right)^3} + {y^4} = {e^{ - t}}\,\,$$ is

A solution of the differential equation $${{{d^2}y} \over {d{x^2}}} - 5{{dy} \over {dx}} + 6y = 0\,$$ is given by

The following differential equation has $$3{{{d^2}y} \over {d{t^2}}} + 4{\left( {{{dy} \over {dt}}} \right)^3} + {y^2} + 2 = x$$