The solution of the initial value problem e dy - dx - x -1- y 0-3 is-A- y-x-1 log x-1-4B- y-x-1 log x-1-3C- y-x-1 log x-1-1D- y-x-1 log x-1-3

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Parts of a linear equation

The solution ...

Question

The solution of the initial value problem e^dy/dx = x + 1, y(0) = 3 is:

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y = (x - 1) \log (x - 1) - (x + 1) \log (x + 1)

\frac{d y}{d c} = \log (\frac{x - 1}{1 + x})

$lim x \to 1 \frac{l o g x}{x - 1}$

lim x \to 1 \frac{l o g x}{x - 1} =

x

{(x + 1)}^{log (x + 1)} = 100 (x + 1)

(y (1 + x^{- 1}) + s i n y) d x + (x + l o g x + x c o s y) d y = 0

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