Quadratic equation with several variables

Question

How does $$y^{2} - 4y -t^{2} - C = 0$$

Become $$y = 2 \pm \sqrt{t^{2} +2C + 4}$$

I know its the quadratic formula but I dont know how it got it that point

The original equation is $$\frac{dy}{dt} = \frac{t}{y-2}$$.

Answer 1

Complete the square in $y$:$$y^2-4y-t^2=C\\y^2-4y+4=t^2+C+4\\(y-2)^2=t^2+C\\y=2\pm\sqrt{t^2+C}$$

Note we can absorb the $4$ into our arbitrary constant $C$ which parameterizes our family of solutions.

From the original equation, notice: $$(y-2)\frac{dy}{dt}=t$$ Let $z=y-2$ so $dz=dy$ giving $$z\frac{dz}{dt}=t\\2z\frac{dz}{dt}=2t\\\int 2z\frac{dz}{dt}\,dt=\int 2t\, dt\\z^2=t^2+C\\z=\pm\sqrt{t^2+C}\\y-2=\pm\sqrt{t^2+C}\\y=2\pm\sqrt{t^2+C}$$

Answer 2

The equation $$\frac{dy}{dt} = \frac{t}{y-2}$$ can be seen in the form \begin{align} (y-2) \, dy = t \, dt \end{align} and upon integration of both sides becomes \begin{align} \frac{y^{2}}{2} - 2 y = \frac{t^{2}}{2} + \frac{c_{0}}{2} \end{align} or $y^{2} - 4 y -(t^{2} + c_{0}) = 0$. Solving this quadratic equation the value of $y$ is of the form \begin{align} y(t) = 2 \pm \sqrt{t^{2} + 4 + c_{0}}. \end{align}