y′′−y′−6y=e−x. (1)
Find a particular solution of (1).
Particular integral is of the form y=Ce−x.
This solution must satisfy the differential equation (1):
(Ce−x)′′−(Ce−x)′−6Ce−x=e−x
Ce−x+Ce−x−6Ce−x=e−x
2C−6C=1,C=−1/4.
Then the particular solution of (1) is
y=−41e−x.
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