It's easy enough to check it yourself isn't it? you have the differential equations \(\displaystyle y_1'= y_1+ 4y_2- 2cos(t)\) and \(\displaystyle y_2'= y_1+ y_2- cos(t)- sin(t)\) and you assert that \(\displaystyle y_1= -2k_1e^{-t}+ 2k_2e^{3t}- sin(t)+ cos(t)\) and \(\displaystyle y_2= k_1e^{-t}+ k_2e^{3t}\).
So if you are correct, \(\displaystyle y_1'= 2k_1e^{-t}+ 6k_2e^{3t}- sin(t)- cos(t)\) while \(\displaystyle y_1+ 4y_2- 2cos(t)=(-2k_1e^{-t}+ 2k_2e^{3t}- sin(t)+ cos(t))+ (4k_1e^{-t}+ 4k_2e^{3t})- 2cos(t)= 2k_1e^{-t}+ 6k_2e^{3t}- sin(t)- cos(t)\).
Yes, those are the same.
Also, \(\displaystyle y_2'= -k_1e^{-t}+ 3k_2e^{3t}\) while \(\displaystyle y_1+ y_2- cos(t)- sin(t)= (-2k_1e^{-t}+ 2k_2e^{3t}- sin(t)+ cos(t))+ (k_1e^{-t}+ k_2e^{3t})- cos(t)- sin(t)= -k_1e^{-t}+ 3k_2e^{3t}- 2sin(t)\).
NO, those are not the same so this is not a correct solution.
Here's a tip: once you have found \(\displaystyle y_1\), do NOT go back and re-solve the whole problem for \(\displaystyle y_2\). Instead, use the fact that \(\displaystyle y_1'= y_1+ 4y_2- 2cos(t)\) so that \(\displaystyle 4y_2= y_1'- y_1+ 2cos(t)\) and \(\displaystyle y_2= \frac{1}{4}y_1'- \frac{1}{4}y_1+ \frac{1}{2}cos(t)\).
