The general solution of the differential equation $$\,\,{{dy} \over {dx}} = \cos \left( {x + y} \right),\,\,$$ with $$c$$ as a constant, is

Consider two solutions $$\,x\left( t \right)\,\,\,\, = \,\,\,{x_1}\left( t \right)\,\,$$ and $$x\left( t \right)\,\,\,\, = \,\,\,{x_2}\left( t \right)\,\,$$ of the differential equation
$$\,\,{{{d^2}x\left( t \right)} \over {d{t^2}}} + x\left( t \right) = 0,t > 0,\,\,$$ such that
$$\,{x_1}\left( 0 \right) = 1,{\left. {{{d{x_1}\left( t \right)} \over {dt}}} \right|_{t = 0}} = 0,$$ $$\,\,\,\,{x_2}\left( 0 \right) = 0,{\left. {{{d{x_2}\left( t \right)} \over {dt}}} \right|_{t = 0}} = 1$$

The wronskian $$\,w\left( t \right) = \left| {{{\matrix{ {{x_1}\left( t \right)} \cr {d{x_1}\left( t \right)} \cr } } \over {dt}}} \right.\left. {{{\matrix{ {{x_2}\left( t \right)} \cr {d{x_2}\left( t \right)} \cr } } \over {dt}}} \right|$$ at $$\,\,t = \pi /2$$

The solution to the differential equation $$\,{{{d^2}u} \over {d{x^2}}} - k{{du} \over {dx}} = 0\,\,\,$$ where $$'k'$$ is a constant, subjected to the boundary conditions $$\,\,u\left( 0 \right) = 0\,\,$$ and $$\,\,\,u\left( L \right) = U,\,\,$$ is

Consider the differential equation $$\,\,{x^2}{{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}} - 4y = 0\,\,\,$$ with the boundary conditions of $$\,\,y\left( 0 \right) = 0\,\,\,$$ and $$\,\,y\left( 1 \right) = 1.\,\,\,$$ The complete solution of the differential equation is