solve the initial value problem dy/dt = ((-2*y)/t) with y(3)=3

To solve the initial value problem $\frac{dy}{dt} = \frac{-2y}{t}$ with $y(3) = 3$, follow these steps:

1. Separate the variables: $\frac{dy}{y} = \frac{-2}{t} dt$

2. Integrate both sides: $\int \frac{1}{y} dy = \int \frac{-2}{t} dt$

   The left side integrates to $\ln|y|$, and the right side integrates to $-2 \ln|t|$: $\ln|y| = -2 \ln|t| + C$

3. Simplify the equation: $\ln|y| = \ln|t^{-2}| + C$

   Exponentiate both sides to remove the natural logarithm: $|y| = e^C \cdot t^{-2}$

   Let $e^C = k$, where $k$ is a constant: $|y| = k t^{-2}$

4. Determine the constant $k$ using the initial condition $y(3) = 3$: $3 = k \cdot 3^{-2}$ $3 = k \cdot \frac{1}{9}$ $k = 27$

5. Write the general solution: $y = 27 t^{-2}$

   Since $y$ is positive and the initial condition does not specify a negative value, we can drop the absolute value.

Therefore, the solution to the initial value problem is: $y(t) = \frac{27}{t^2}$