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