1. **State the problem:** Solve the first-order linear differential equation $$y' - \frac{3}{t}y = \frac{\cos t}{t^{-3}}, \quad y(\pi) = 0, \quad t > 0.$$ 2. **Rewrite the equation:** Note that $\frac{\cos t}{t^{-3}} = \cos t \cdot t^3 = t^3 \cos t$. So the equation becomes $$y' - \frac{3}{t}y = t^3 \cos t.$$ 3. **Identify the integrating factor:** The standard form is $$y' + p(t)y = q(t),$$ where here $p(t) = -\frac{3}{t}$ and $q(t) = t^3 \cos t$. The integrating factor $\mu(t)$ is $$\mu(t) = e^{\int p(t) dt} = e^{\int -\frac{3}{t} dt} = e^{-3 \ln t} = e^{\ln t^{-3}} = t^{-3}.$$ 4. **Multiply both sides by the integrating factor:** $$t^{-3} y' - t^{-3} \frac{3}{t} y = t^{-3} \cdot t^3 \cos t,$$ which simplifies to $$t^{-3} y' - 3 t^{-4} y = \cos t.$$ 5. **Recognize the left side as a derivative:** $$\frac{d}{dt} \left(t^{-3} y \right) = \cos t.$$ 6. **Integrate both sides:** $$\int \frac{d}{dt} \left(t^{-3} y \right) dt = \int \cos t dt,$$ which gives $$t^{-3} y = \sin t + C.$$ 7. **Solve for $y$:** $$y = t^{3} (\sin t + C).$$ 8. **Apply the initial condition $y(\pi) = 0$:** $$0 = \pi^{3} (\sin \pi + C) = \pi^{3} (0 + C) = \pi^{3} C,$$ so $$C = 0.$$ 9. **Final solution:** $$\boxed{y(t) = t^{3} \sin t}.$$