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

If $$\,y = f\left( x \right)\,\,$$ is the solution of $${{{d^2}y} \over {d{x^2}}} = 0$$ with the boundary conditions $$y=5$$ at $$x=0,$$ and $$\,{{dy} \over {dx}} = 2$$ at $$x=10,$$ $$f(15)=$$_______.

The matrix form of the linear system $${{dx} \over {dt}} = 3x - 5y$$ and $$\,{{dy} \over {dt}} = 4x + 8y\,\,$$ 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 $$\,\,t = \pi /2$$