The particular solution of the initial value problem given below is
$$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + 12\frac{{{\text{dy}}}}{{{\text{dx}}}} + 36{\text{y}} = 0$$     with y(0) = 3 and $${\left. {\frac{{{\text{dy}}}}{{{\text{dx}}}}} \right|_{{\text{x}} = 0}} = - 36$$

The general solution of the differential equation, $$\frac{{{{\text{d}}^4}{\text{y}}}}{{{\text{d}}{{\text{x}}^4}}} - 2\frac{{{{\text{d}}^3}{\text{y}}}}{{{\text{d}}{{\text{x}}^3}}} + 2\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - 2\frac{{{\text{dy}}}}{{{\text{dx}}}} + {\text{y}} = 0$$       is

The solution of the differential equation, $${\text{y}}\sqrt {1 - {{\text{x}}^2}} {\text{dy}} + {\text{x}}\sqrt {1 - {{\text{y}}^2}} {\text{dx}} = {\text{0}}$$      is

The maximum value of the solution y(t) of the differential equation \[{\rm{y}}\left( {\rm{t}} \right) + {\rm{\ddot y}}\left( {\rm{t}} \right) = 0\]    with initial conditions \[{\rm{\dot y}}\left( 0 \right) = 1\]  and y(0) = 1, for t ≥ 0 is

The order of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{t}}^2}}} + {\left( {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right)^3} + {{\text{y}}^4} = {{\text{e}}^{ - {\text{t}}}}$$     is