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Special solutions of differential equation y '- y = 2xe ^ (2x), y (0) = 1
The first solution is y '- y = 0
dy/y=dx
ln|y|=x+C
y=Ce^x
Let y = UE ^ X by constant variation method
y'=u'e^x+ue^x
y'-y=u'e^x=2xe^(2x)
That is u '= 2xe ^ X
We get u = 2xe ^ x-2e ^ x + C
So y = (2xe ^ x-2e ^ x + C) e ^ x
Substituting x = 0, y = 1, C = 3
So y = (2xe ^ x-2e ^ x + 3) e ^ x
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