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प्रश्न
Solve the differential equation `(tan^(-1) x- y) dx = (1 + x^2) dy`
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उत्तर
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संबंधित प्रश्न
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Solution: The equation `("d"y)/("d"x) - y` = 2x
is of the form `("d"y)/("d"x) + "P"y` = Q
where P = `square` and Q = `square`
∴ I.F. = `"e"^(int-"d"x)` = e–x
∴ the solution of the linear differential equation is
ye–x = `int 2x*"e"^-x "d"x + "c"`
∴ ye–x = `2int x*"e"^-x "d"x + "c"`
= `2{x int"e"^-x "d"x - int square "d"x* "d"/("d"x) square"d"x} + "c"`
= `2{x ("e"^-x)/(-1) - int ("e"^-x)/(-1)*1"d"x} + "c"`
∴ ye–x = `-2x*"e"^-x + 2int"e"^-x "d"x + "c"`
∴ e–xy = `-2x*"e"^-x+ 2 square + "c"`
∴ `y + square + square` = cex is the required general solution of the given differential equation
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Solve:
`xsinx dy/dx + (xcosx + sinx)y` = sin x
