A curve passes through (2, 0) and the slope of tangent at a point P(x, y) is equal to . Then equation of the curve is

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= (x + 1) +
Putting x + 1 = X, y - 3 = Y,
the equation becomes
. Y = X [L.D.E.]
I.F. = e^{∫(-1/ X)dX} = e^{-log x} =
∴ the solution is
Y. = c + ∫ X dx = c + X
or

= c + x + 1
x = 2, y = 0 ⇒ = c + 2 + 1
⇒ c = - 4
∴ the quation of the curve is
= x - 3 or y = x² - 2x.

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