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 = Aem1t + Bem2t
Therefore we have the GS to our 2nd ODE given above to be: y = Ae-t + Be-2t
IG
Answered by Isaac G. • Further Mathematics tutor
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