# 3 \\( \\frac { d y } { d t } = - \\frac { 2 t } { y } \\), \\( y = - 6 \\) when \\( t = 0…

Explanation:

Step1: Separate variables

We have the differential equation (\frac{dy}{dt}=-\frac{2t}{y}). Separate the variables: (y;dy=- 2t;dt).

Step2: Integrate both sides

Integrate (\int y;dy=\int - 2t;dt). Using the power - rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we get (\frac{y^{2}}{2}=-t^{2}+C).

Step3: Use the initial condition

Given (y=-6) when (t = 0). Substitute (t = 0) and (y=-6) into (\frac{y^{2}}{2}=-t^{2}+C). (\frac{(-6)^{2}}{2}=-0^{2}+C), so (C = 18). The particular solution is (\frac{y^{2}}{2}=-t^{2}+18), or (y^{2}=-2t^{2}+36).

Step4: Find (t) when (y=-3\sqrt{2})

Substitute (y=-3\sqrt{2}) into (y^{2}=-2t^{2}+36). ((-3\sqrt{2})^{2}=-2t^{2}+36). Since ((-3\sqrt{2})^{2}=18), we have the equation (18=-2t^{2}+36). Rearrange it: (2t^{2}=36 - 18), (2t^{2}=18), (t^{2}=9). Take the square root of both sides: (t=\pm3).

Answer:

The particular solution is (y^{2}=-2t^{2}+36). When (y=-3\sqrt{2}), (t = 3) or (t=-3).