Question

Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve at any point (x, y) is `(y - 1)/(x^2 + x)`

Answer 1

Given that the slope of the tangent to the curve at (x, y) is `("d"y)/("d"x) = (y - 1)/(x^2 + x)`

⇒ `("d"y)/(y - 1) = ("d"x)/(x^2 + x)`

Integrating both sides, we have

`int ("d"y)/(y - 1) = int ("d"x)/(x^2 + x)`

⇒ `int ("d"y)/(y - 1) = int ("d"x)/(x^2 + x + 1/4 - 1/4)`  ...[making perfect square]

⇒ `int ("d"y)/(y - 1) = int ("d"x)/((x + 1/2)^2 - (1/2)^2`

⇒ `log|y - 1| = 1/(2 xx 1/2) log|(x + 1/2 - 1/2)/(x + 1/2 + 1/2)|`

⇒ `log|y - 1| = log|x/(x + 1)| + log "c"`

⇒ `log|y - 1| = log|"c"(x/(x + 1))|`

∴ y – 1 = `("c"x)/(x + 1)` 

⇒ `(y - 1)(x + 1)` = cx

Since, the line is passing through the point (1, 0), then (0 – 1) (1 + 1) = c(1)

⇒ c = 2

Hence, the required solution is (y – 1)(x + 1) = 2x.