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**Solve the following differential equation:**

`(1 - "x"^2) "dy"/"dx" + "2xy" = "x"(1 - "x"^2)^(1/2)`

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#### Solution

`(1 - "x"^2) "dy"/"dx" + "2xy" = "x"(1 - "x"^2)^(1/2)`

∴ `"dy"/"dx" + ("2x"/(1 - "x"^2))"y" = "x"/(1 - "x"^2)^(1/2)`

This is the linear differential equation of the form

`"dy"/"dx" + "P" * "y" = "Q",` where P = `"2x"/(1 - "x"^2)` and Q = `"x"/(1 - "x"^2)^(1/2)`

∴ I.F. = `"e"^(int "P dx") = "e"^(int "2x"/"1 - x"^2"dx")`

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

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

∴ the solution of (1) is given by

`"y"*("I.F.") = int "Q" * ("I.F.") "dx" + "c"`

∴ `"y" * 1/(1 - "x"^2) = int "x"/(1 - "x")^(1/2) * 1/(1 - "x"^2)` dx + c

∴ `"y"/((1 - "x"^2)) = int "x"/(1 - "x"^2)^(3/2) "dx" + "c"`

Put 1 - x^{2} = t

∴ x dx = - `"dt"/2`

∴ `"y"/(1 - "x"^2) = int 1/"t"^(3/2) * (- "dt")/2 + "c"`

∴ `"y"/(1 - "x"^2) = - 1/2 int "t"^(- 3/2) "dt" + "c"`

∴ `"y"/(1 - "x"^2) = - 1/2 * "t"^(- 1/2)/(- 1/2) + "c"`

∴ `"y"/(1 - "x"^2) = 1/(1 - "x"^2)^(1/2) + "c"`

∴ y = `sqrt(1 - "x"^2) + "c" (1 - "x"^2)`

This is the general solution.