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13-03-2024
Mathematics

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y(3)−2y′′+y′=7+5xeˣ; y(0)=y′(0)=0, y′′(0)=1. The solution is y(x)=
a)7x+5xeˣ+x³
b)5x+7xeˣ+x³
c)7+5xeˣ+x³
d)7x+xeˣ+x³

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