1. **State the problem:** Solve the initial value problem $$ty' + 4y = t^2 - t + 5, \quad y(1) = 7, \quad t > 0.$$ 2. **Rewrite the equation:** Divide both sides by $t$ to get the standard linear form: $$y' + \frac{4}{t}y = t - 1 + \frac{5}{t}.$$ 3. **Identify integrating factor:** The integrating factor $\mu(t)$ is given by $$\mu(t) = e^{\int \frac{4}{t} dt} = e^{4 \ln t} = t^4.$$ 4. **Multiply entire equation by $\mu(t)$:** $$t^4 y' + 4 t^3 y = t^4 \left(t - 1 + \frac{5}{t}\right) = t^5 - t^4 + 5 t^3.$$ 5. **Recognize left side as derivative:** $$\frac{d}{dt}(t^4 y) = t^5 - t^4 + 5 t^3.$$ 6. **Integrate both sides:** $$t^4 y = \int (t^5 - t^4 + 5 t^3) dt = \frac{t^6}{6} - \frac{t^5}{5} + \frac{5 t^4}{4} + C.$$ 7. **Solve for $y$:** $$y = \frac{1}{t^4} \left( \frac{t^6}{6} - \frac{t^5}{5} + \frac{5 t^4}{4} + C \right) = \frac{t^2}{6} - \frac{t}{5} + \frac{5}{4} + \frac{C}{t^4}.$$ 8. **Apply initial condition $y(1) = 7$:** $$7 = \frac{1}{6} - \frac{1}{5} + \frac{5}{4} + C \implies 7 = \frac{1}{6} - \frac{1}{5} + \frac{5}{4} + C.$$ Calculate constants: $$\frac{1}{6} - \frac{1}{5} = \frac{5 - 6}{30} = -\frac{1}{30}, \quad -\frac{1}{30} + \frac{5}{4} = -\frac{1}{30} + \frac{37.5}{30} = \frac{36.5}{30} = \frac{73}{60}.$$ So, $$7 = \frac{73}{60} + C \implies C = 7 - \frac{73}{60} = \frac{420}{60} - \frac{73}{60} = \frac{347}{60}.$$ 9. **Final solution:** $$\boxed{y = \frac{t^2}{6} - \frac{t}{5} + \frac{5}{4} + \frac{347}{60 t^4}}.$$