Find the GS to the following 2nd ODE: d^2y/dx^2 + 3(dy/dx) + 2 = 0

Set up the auxiliary equation by letting (dy/dx) = m
So we have: m² + 3m + 2 = 0
Solve for m and we get: (m+1)(m+2) = 0Therefore, m₁=-1 and m₂=-2
Now we see we have 2 different real numbers as the solutions to our auxiliary equation. So employ the GS in the form of: y = Ae^m1t + Be^m2t
Therefore we have the GS to our 2nd ODE given above to be: y = Ae^-t + Be^-2t