the given equation can be written as
"\\frac{1}{\\sqrt{y}} \\frac{dy}{dx} + \\frac{2x\\sqrt{y}}{2(1-x^2)} = x" .....(1)
this is of the form f'(y) y' + P f(y) = Q
where P = "\\frac{x}{2(1-x^2)}"
Q= x
f(y) = 2 "\\sqrt{y}"
let v= 2"\\sqrt{y}"
differentiating,
"\\frac{1}{\\sqrt{y}} \\frac{dy}{dx} = \\frac{dv}{dx}"
(1) becomes
"\\frac{dv}{dx} + \\frac{xv}{2(1-x^2)} = x"
this is of the form "\\frac{dv}{dx} + Av = B"
"A = \\frac{x}{2(1-x^2)} and B=x"
integrating factor = "e^{\\intop Pdx} = e^{\\intop \\frac{x}{2(1-x^2)}dx} = (1-x^2)^{(-1\/4)}"
the general solution is given by:
v (1-x2)-1/4 = "\\int" x. (1-x2)-1/4 dx + c
2"\\sqrt{y}" = -2/3 (1-x2) + c(1-x2)1/4
now, given y(0)=1
2 = -2/3 + c
c= 8/3
so, the solution is
2"\\sqrt{y}" = -2/3 (1-x2) + 8/3(1-x2)1/4
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