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प्रश्न
Find the general solution of the differential equation `(1 + y^2) + (x - "e"^(tan - 1y)) "dy"/"dx"` = 0.
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उत्तर
Given equation is `(1 + y^2) + (x - "e"^(tan^(-1) y)) "dy"/"dx"` = 0
⇒ `(x - "e"^(tan^-1y)) "dy"/"dx" = -(1 + y^2)`
⇒ `"dy"/"dx" = (-(1 + y^2))/(x - "e"^(tan^-1 y))`
⇒ `"dx"/"dy" = (x - "e"^(tan^-1y))/(-(1 + y^2))`
⇒ `"dx"/"dy" = - x/((1 + y^2)) + ("e"^(tan^-1y))/(1 + y^2)`
⇒ `"dx"/"dy" + x/((1 + y^2)) = ("e"^(tan^-1 y))/(1 + y^2)`
Here, P = `1/(1 + y^2)` and Q = `("e"^(tan^-1 y))/(1 + y^2)`
∴ Integrating factor I.F. = `"e"^(int Pdy)`
= `"e"^(int 1/(1 + y^2) "d"y)`
= `"e"^(tan^-1 y)`
∴ Solution is `x . "I"."F". = int "Q". "I"."F". "d"y + "c"`
⇒ `x . "e"^(tan^-1 y) = int ("e"^(tan^-1 y))/(1 + y^2) * "e"^(tan^-1 y) "dy" + "c"`
Put `"e"^(tan^-1 y)` = t
∴ `"e"^(tan^-1 y) * 1/(1 + y^2) "dy"` = dt
∴ `x . "e"^(tan^-1 y) = int "t" . "dt" + "c"`
⇒ `x . "e"^(tan^-1 y) = 1/2 "t"^2 + "c"`
⇒ `x . "e"^(tan^-1 y) = 1/2 ("e"^(tan^-1 y))^2 + "c"`
⇒ x = `1/2 ("e"^(tan^-1 y)) + "c"/("e"^(tan^-1 y))`
⇒ 2x = `"e"^(tan^-1 y) + (2"c")/("e"^(tan^-1 y)`
⇒ `2x . "e"^(tan^-1 y) = ("e"^(tan^-1y))^2 + 2"c"`
Hence, this is the required general solution.
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