1. **State the problem:** We are given the differential equation $$ty' + 2y = 4t^2$$ with the initial condition $$y(1) = 2$$. 2. **Check the proposed solution:** The proposed solution is $$y = t^2 + \frac{1}{t^2}$$. 3. **Find the derivative of the proposed solution:** $$y' = \frac{d}{dt}\left(t^2 + \frac{1}{t^2}\right) = 2t - 2t^{-3}$$. 4. **Substitute $y$ and $y'$ into the differential equation:** $$t y' + 2y = t(2t - 2t^{-3}) + 2\left(t^2 + \frac{1}{t^2}\right)$$ 5. **Simplify the expression:** $$= 2t^2 - 2t^{-2} + 2t^2 + 2t^{-2}$$ 6. **Cancel terms:** $$= (2t^2 + 2t^2) + (-2t^{-2} + 2t^{-2}) = 4t^2 + 0 = 4t^2$$ 7. **Verify initial condition:** $$y(1) = 1^2 + \frac{1}{1^2} = 1 + 1 = 2$$ which matches the given initial condition. **Conclusion:** The proposed solution satisfies both the differential equation and the initial condition. **Final answer:** The solution $$y = t^2 + \frac{1}{t^2}$$ is correct.