1. **State the problem:** Solve for $y(t)$ given the differential equation $$(100 - t)^{-5} y(t) = 8 \int (100 - t)^{-5} \, dt.$$\n\n2. **Rewrite the integral:** We need to evaluate $$\int (100 - t)^{-5} \, dt.$$\n\n3. **Use substitution:** Let $$u = 100 - t,$$ so $$du = -dt,$$ or $$dt = -du.$$\n\n4. **Rewrite the integral in terms of $u$:** $$\int u^{-5} (-du) = - \int u^{-5} \, du.$$\n\n5. **Integrate:** $$- \int u^{-5} \, du = - \frac{u^{-4}}{-4} + C = \frac{u^{-4}}{4} + C = \frac{(100 - t)^{-4}}{4} + C.$$\n\n6. **Substitute back into the original equation:** $$(100 - t)^{-5} y(t) = 8 \times \frac{(100 - t)^{-4}}{4} + 8C = 2 (100 - t)^{-4} + 8C.$$\n\n7. **Solve for $y(t)$:** Multiply both sides by $(100 - t)^5$ to isolate $y(t)$:\n$$y(t) = (100 - t)^5 \left[ 2 (100 - t)^{-4} + 8C \right] = 2 (100 - t) + 8C (100 - t)^5.$$\n\n8. **Simplify the constant:** Let $$C' = 8C,$$ so the general solution is $$y(t) = 2 (100 - t) + C' (100 - t)^5.$$\n\n**Final answer:** $$\boxed{y(t) = 2 (100 - t) + C (100 - t)^5}.$$