The figure shows the plot of $$y$$ as a function of $$x$$

The function shown in the solution of the differential equation (assuming all initial conditions to be zero) is

The type of the partial differential equation $${{\partial f} \over {\partial t}} = {{{\partial ^2}f} \over {\partial {x^2}}}\,\,is$$

With initial condition $$x\left( 1 \right)\,\,\, = \,\,\,\,0.5,\,\,\,$$ the solution of the differential equation, $$\,\,\,t{{dx} \over {dt}} + x = t\,\,\,$$ is

Consider the differential equation $${{dy} \over {dx}} + y = {e^x}$$ with $$y(0)=1.$$ Then the value of $$y(1)$$ is