so the problem at hand is:
$(2x-3)\frac{dx}{dt}+t-2=0$
and the solution is
$x(t)=\frac{3}{2}\frac{\sqrt{C-2t(t-4)+9}}{2}$
>
Originally. I tried the exactness test but found that both $\frac{\partial p}{\partial t}$ and $\frac{\partial q}{\partial x}$ equalled zero... so I tried another route and made some progress... I rearranged the equation so that:
$(2x-3)dx=-(t-2)dt$
$\int{(2x-3)dx}=-\int{(t-2)dt}$
$x^{2}-3x=-\frac{t^{2}}{2}+2t+C$
>
now, using the solution as a guide, I got my solution so far to:
$2x^{2}-6x=-t(t-4)+D$
This looks abit more promising, but I still get stuck... I figured that this isnt just a problem with algebraic manipulation, I'm actually missing something here or doing something wrong... Any ideas??
$$x^{2}-3x=-\frac{t^{2}}{2}+2t+C\\(x-3/2)^2-\frac94=-\frac{t^2}2+2t+C\\x=\frac32\pm\sqrt{\frac{-t^2+4t+C}{2}}\\x=\frac32\pm\frac{\sqrt{C-2t(t-4)+9}}2$$ At some steps, I changed the integration constant a little bit, this doesn't make any difference since the constant is arbitrary anyway.
Also, this slightly disagrees with what you give as the solution. I guess your question must have a typo?