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Find the general solution of y"'-y"-y'+y=2e-t+3.
Here's the work:
r3-r2-r+1=r2(r-1)-(r-1)=(r-1)(r2-1)=(r2-1)2(r+1)
r=1, -1
y=c1et+c2tet+c3e-t
The answer in the textbook is y=c1et+c2tet+c3e-t+(1/2)te-t+3 but I don't know how to get the last 2 terms. Help me...
Here's the work:
r3-r2-r+1=r2(r-1)-(r-1)=(r-1)(r2-1)=(r2-1)2(r+1)
r=1, -1
y=c1et+c2tet+c3e-t
The answer in the textbook is y=c1et+c2tet+c3e-t+(1/2)te-t+3 but I don't know how to get the last 2 terms. Help me...