With initial condition $$x\left( 1 \right)\,\,\, = \,\,\,\,0.5,\,\,\,$$ the solution of the differential equation, $$\,\,\,t{{dx} \over {dt}} + x = t\,\,\,$$ is

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

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