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

Consider the following differential equation $${{dy} \over {dt}} = - 5y;$$ initial condition: $$y=2$$ at $$t=0.$$
The value of $$y$$ at $$t=3$$ is

The solution of the initial value problem $$\,\,{{dy} \over {dx}} = - 2xy;y\left( 0 \right) = 2\,\,\,$$ is

The partial differential equation $$\,\,{{\partial u} \over {\partial t}} + u{{\partial u} \over {\partial x}} = {{{\partial ^2}u} \over {\partial {x^2}}}\,\,\,$$ is a