Given the differential equation,
d2ydx2+3dydx−54y=0.......(1).
Let y=cemx for c≠0 be a trial solution of equation (1).
Then the auxiliary equation is
m2+3m−54=0
or, m2+9m−6m−54=0
or, (m+9)(m−6)=0.
or, m=6,−9.
So the solution is y=c1e−9x+c2e6x. [ Where c1 and c2 are arbitrary constant.]