Question

Find the equation of the curve whose slope is `(y - 1)/(x^2 + x)` and which passes through the point (1, 0)

Answer 1

Given the slope of the equation be `(y - 1)/(x^2 + x)`

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

THe equation can be written as

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

Take `1/(x^2 + x) = 1/(x(x + 1)) = "A"/x + "B"/(x + 1)`  .......[Solve by pratical fraction]

`1/(x(x + 1)) = ("A"(x + 1) + "B"(x))/(x(x + 1))`

1 = A(x + 1) + B(x)

Put x = – 1, Put x = 0

1 = A(0) + B(– 1), 1 = A(0 + 1) + B(0)

1 = – B, 1 = A

B = – 1, A = 1

∴ `1/(x^2 + x) = 1/x + 1/(x + 1)` .........(2)

Substituting equation (2) in equation (1), we get

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

Taking integrating on both sides, we get

log(y – 1) = log x – log(x + 1) + log C

log(y – 1) = log C + log x – log(x + 1)

= log Cx – log(x + 1)

log(y – 1) = `log (("Cx")/(x + 1))`

y – 1= `"Cx"/(x + 1)` .........(3)

The curve passes through (1, 0), we get

0 – 1 = `("C"(1))/(1 + 1)`

– 1 = `"C"/2`

– 2 = C

(3) ⇒ y – 1= `- (2x)/(x + 1)`

y = `1 - (2x)/(x + 1)`

= `(x + 1 - 2x)/(x + 1)`

= `(1 - x)/(x + 1)`

∴ y = `(1 - x)/(x + 1)`