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Integrating factor of the differential equation dd(1-x2)dydx-xy = 1 is ______. - Mathematics

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Question

Integrating factor of the differential equation `(1 - x^2) ("d"y)/("d"x) - xy` = 1 is ______.

Options

  • – x

  • `x/(1 + x^2)`

  • `sqrt(1 - x^2)`

  • `1/2 log (1 - x^2)`

MCQ
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Solution

Integrating factor of the differential equation `(1 - x^2) ("d"y)/("d"x) - xy` = 1 is `sqrt(1 - x^2)`.

Explanation:

The given differential equation is `(1 - x^2) ("d"y)/("d"x) - xy` = 1

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

Here, P = `x/(1 - x^2)` and Q = `1/(1 - x^2)`

∴ Integrating factor I.F. = `"e"^(int Pdx)`

= `"e"^(int (-x)/(1 - x^2) dx)`

= `"e"^(1/2 log(1 - x^2))`

= `sqrt(1 - x^2)`

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Chapter 9: Differential Equations - Exercise [Page 197]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 9 Differential Equations
Exercise | Q 47 | Page 197

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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"`

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