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Engineering Maths
until the integrating parts is correct:
ln(y-1) = ln x + B
where B is an arbitrary constant
pls explain why is it so, for the following solution.
On solving this eqn for y, by first taking the exponential of both sides, we obtain
y = 1 ± e^lnx+B
= 1 ± e^B . e^lnx
= ???
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