$(5)$ Verify that the constant functions $x(t)=0,x(t)=2$ and $x(t)=4$ are solutions to the differential equation

$\frac{dx}{dt}=\frac{t{c}^{-x}(x-4)(2x-x^2)}{x-3}$

Now consider the initial value problern

$\frac{dx}{dt}=\frac{t{c}^{-x}(x-4)(2x-x^2)}{x-3}$

$x(3)=\frac{3}{2}$