To solve the given initial value problem:

$(1 - 3t) \frac{dy}{dt} - y = 0, \quad y(2) = -6$

we proceed step by step.

Step 1: Rearranging the Equation

We can rewrite the equation in a form that is easier to handle. Begin by isolating the derivative $\frac{dy}{dt}$:

$(1 - 3t) \frac{dy}{dt} = y$

$\frac{dy}{dt} = \frac{y}{1 - 3t}$

This is a separable differential equation, so we can separate the variables $y$ and $t$.

Step 2: Separating Variables

We rearrange the equation to separate the variables:

$\frac{dy}{y} = \frac{dt}{1 - 3t}$

Step 3: Integrating Both Sides

Now, we integrate both sides:

$\int \frac{1}{y} \, dy = \int \frac{1}{1 - 3t} \, dt$

The integral on the left-hand side is straightforward:

$\ln|y| = \int \frac{1}{1 - 3t} \, dt$

For the integral on the right-hand side, use substitution. Let $u = 1 - 3t$, so $du = -3dt$, and $dt = -\frac{du}{3}$.

$\int \frac{1}{1 - 3t} \, dt = \int \frac{1}{u} \left(-\frac{du}{3}\right) = -\frac{1}{3} \ln|u| + C = -\frac{1}{3} \ln|1 - 3t| + C$

So we have:

$\ln|y| = -\frac{1}{3} \ln|1 - 3t| + C$

Step 4: Solving for $y$

Now exponentiate both sides to eliminate the natural logarithm:

$|y| = e^{-\frac{1}{3} \ln|1 - 3t| + C}$

This simplifies to:

$|y| = e^C \cdot |1 - 3t|^{-\frac{1}{3}}$

Let $A = e^C$, where $A$ is a constant, so:

$|y| = A \cdot |1 - 3t|^{-\frac{1}{3}}$

Thus, the general solution is:

$y = A \cdot (1 - 3t)^{-\frac{1}{3}}$

Step 5: Using the Initial Condition

We are given the initial condition $y(2) = -6$. Substitute $t = 2$ and $y = -6$ into the general solution:

$-6 = A \cdot (1 - 3(2))^{-\frac{1}{3}} = A \cdot (-5)^{-\frac{1}{3}}$

Since $(-5)^{-\frac{1}{3}} = -\frac{1}{5^{1/3}}$, we get:

$-6 = A \cdot \left(-\frac{1}{5^{1/3}}\right)$

Solving for $A$:

$A = 6 \cdot 5^{1/3}$

Thus, the particular solution is:

$y = 6 \cdot 5^{1/3} \cdot (1 - 3t)^{-\frac{1}{3}}$

Final Answer:

The solution to the initial value problem is:

$y = 6 \cdot 5^{1/3} \cdot (1 - 3t)^{-\frac{1}{3}}$

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Would you like further clarification on any part of the solution?

Here are some related questions you could explore:

1. How does the solution change if the differential equation is altered to a non-separable form?
2. What happens if the initial condition is different, say $y(1) = 3$?
3. How do we solve a first-order linear differential equation using an integrating factor?
4. How does the behavior of the solution change as $t$ approaches $\frac{1}{3}$?
5. What would the solution look like if the coefficient of $\frac{dy}{dt}$ were not linear?

Tip: When solving separable differential equations, always be careful about domain restrictions, especially if the function inside the logarithm or a power becomes zero!