1. **State the problem:** Solve the initial value problem (IVP) given by the differential equation $$(1 - 9t) \frac{dy}{dt} - y = 0$$ with the initial condition $$y(3) = -4$$. 2. **Rewrite the equation:** First, isolate $$\frac{dy}{dt}$$: $$ (1 - 9t) \frac{dy}{dt} = y $$ $$ \frac{dy}{dt} = \frac{y}{1 - 9t} $$ 3. **Recognize the type of equation:** This is a separable differential equation. We can write it as: $$ \frac{dy}{y} = \frac{dt}{1 - 9t} $$ 4. **Integrate both sides:** $$ \int \frac{1}{y} \, dy = \int \frac{1}{1 - 9t} \, dt $$ 5. **Integrate the left side:** $$ \int \frac{1}{y} \, dy = \ln|y| + C_1 $$ 6. **Integrate the right side:** Use substitution $$u = 1 - 9t$$, so $$du = -9 dt$$, thus $$dt = -\frac{1}{9} du$$: $$ \int \frac{1}{u} \left(-\frac{1}{9}\right) du = -\frac{1}{9} \int \frac{1}{u} du = -\frac{1}{9} \ln|u| + C_2 = -\frac{1}{9} \ln|1 - 9t| + C_2 $$ 7. **Combine constants:** Let $$C = C_2 - C_1$$, then $$ \ln|y| = -\frac{1}{9} \ln|1 - 9t| + C $$ 8. **Exponentiate both sides:** $$ |y| = e^{C} \cdot |1 - 9t|^{-\frac{1}{9}} $$ Let $$A = e^{C} > 0$$, so $$ y = A (1 - 9t)^{-\frac{1}{9}} $$ 9. **Apply initial condition:** Given $$y(3) = -4$$, $$ -4 = A (1 - 9 \cdot 3)^{-\frac{1}{9}} = A (1 - 27)^{-\frac{1}{9}} = A (-26)^{-\frac{1}{9}} $$ 10. **Solve for $$A$$:** $$ A = -4 \times (-26)^{\frac{1}{9}} $$ 11. **Write the final solution:** $$ y(t) = -4 (-26)^{\frac{1}{9}} (1 - 9t)^{-\frac{1}{9}} $$ This is the exact solution to the initial value problem.