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प्रश्न
Solve the following Linear differential equation:
`(y - "e"^(sin^-1)x) ("d"x)/("d"y) + sqrt(1 - x^2)` = 0
उत्तर
`(y - "e"^(sin^-1x)) ("d"x)/("d"y) + sqrt(1 - x^2)` = 0
`(y - "e"^(sin^-1x)) ("d"x)/("d"y) = - sqrt(1 - x^2)`
`(y - "e"^(sin^-1x)) = - sqrt(1 - x^2) ("d"y)/("d"x)`
`sqrt(1 - x^2) ("d"y)/("d"x) + y = "e"^(sin^-1)x`
÷ By `sqrt(1 - x^2) ("d"y)/("d"x) + 1/sqrt(1 - x^2) y = ("e"^(sin^-1)x)/sqrt(1 - x^2)`
Thus, the given differential equation is Linear
P = `1/sqrt(1 - x^2)`
Q = `("e"^(sin^-1)x)/sqrt(1 - x^2)`
I.F = `"e"^(int "Pd"x)`
= `"e" ^(int 1/sqrt(1 - x^2) "d"x)`
= `"e"^(sin^-1x)`
So, the required solution is
`y xx "I.F" = int "Q" xx "I.F" "d"x + "c"`
`y "e"^(sin^-1x) = int ("e"^(sin^-1x))/sqrt(1 - x^2) + "e"^(sin^-1x) "d"x + "c"`
t = `sin^-1x`
dt = `1/sqrt(1 - x^2) "d"x`
`y "e"^(sin^-1x) = int "e"^"t" * "e"^"t" "dt" + "c"`
= `int "e"^(2"t") "dt" + "c"`
`y "e"^(sin^-1x) = "e"^(2"t")/2 + "c"`
∴ `y "e"^(sin^-1x) = ("e"^(2 sin^-1x))/2 + "c"`
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