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Question
tan–1x + tan–1y = c is the general solution of the differential equation ______.
Options
`("d"y)/("d"x) = (1 + y^2)/(1 + x^2)`
`("d"y)/("d"x) = (1 + x^2)/(1 + y^2)`
(1 + x2)dy + (1 + y2)dx = 0
(1 + x2)dx + (1 + y2)dy = 0
Solution
tan–1x + tan–1y = c is the general solution of the differential equation (1 + x2)dy + (1 + y2)dx = 0.
Explanation:
Given equation is tan–1x + tan–1y = c
Differentiating w.r.t. x, we have
`1/(1 + x^2) + 1/(1 + y^2) * ("d"y)/("d"x)` = 0
⇒ `(1/(1 + y^2)) ("d"y)/("d"x) = -(1/(1 + x^2))`
⇒ `("d"y)/("d"x) = -((1 + y^2)/(1 + x^2))`
⇒ (1 + x2)dy = – (1 + y2)dx
⇒ (1 + x2)dy + (1 + y2)dx = 0.
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