# Solve the following differential equation: xdydx2xyxx(1-x2)dydx+2xy=x(1-x2)12 - Mathematics and Statistics

Sum

Solve the following differential equation:

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

#### 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 - x2 = 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.

Concept: Methods of Solving First Order, First Degree Differential Equations - Linear Differential Equations
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