Differential EquationHard

Question

If = 1 + x + y + xy and y (- 1) = 0, then function y is

Options

A.e(1-x)2/2
B.e(1-x)2/2 -1
C.loge (1 + x) - 1
D.1 + x

Solution

= 1 + x + y + xy  = (1 + x) (1 + y)
⇒ = ∫(1 + x)dx
⇒ ln(1 + y) = x + + c
y(-1) = 0 ⇒ c =
ln(1 + y) = x +
⇒ y = e - 1

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