Question

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

particular solution:
to advance in the circuit, find \\( t \\) when \\( y = - 3 \sqrt { 2 } \\).

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 \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), 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 \(t=-3\).