Solution of ∇²T = 0 in a square domain (0 < x < 1 and 0 < y < 1) with boundary conditions:

T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1, y) = 1 + y is

The differential equation $$\,{{{d^2}y} \over {d{x^2}}} + 16y = 0$$ for $$y(x)$$ with the two boundary conditions $${\left. {{{dy} \over {dx}}} \right|_{x = 0}} = 1$$ and $${\left. {{{dy} \over {dx}}} \right|_{x = {\pi \over 2}}} = - 1$$ has

Consider the following partial differential equation for $$u(x,y)$$ with the constant $$c>1:$$ $$\,{{\partial u} \over {\partial y}} + c{{\partial u} \over {dx}} = 0\,\,$$ solution of this equation is

Find the solution of $${{{d^2}y} \over {d{x^2}}} = y$$ which passes through origin and the point $$\left( {ln2,{3 \over 4}} \right)$$