What is the solution for (1-x^2)y^'-xy=1 ?

The solution for (1-x^2)y^'-xy=1 is y=(arcsin(x))/(sqrt(1-x^2))+(c_{1})/(sqrt(1-x^2))

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(1-x^2)y^'-xy=1

(1 - x^{2}) y^{^{'}} - x y = 1

y = \frac{arcsin ( x )}{1 - x ^{2}} + \frac{c _{1}}{1 - x ^{2}}

Solution steps

(1 - x^{2}) y^{'} (x) - x y = 1

Solve linear ODE:

y = \frac{arcsin ( x )}{1 - x ^{2}} + \frac{c _{1}}{1 - x ^{2}}

y = \frac{arcsin ( x )}{1 - x ^{2}} + \frac{c _{1}}{1 - x ^{2}}

\frac{2 x}{x ^{2} - 7}

y^{^{'}} + 6 y = 3

\frac{d}{d x} (cos (x) - cos (2 x))

f (x) = \frac{x}{x + 1}

\int 3 x^{2} - x^{3} d x

What is the solution for (1-x^2)y^'-xy=1 ?
The solution for (1-x^2)y^'-xy=1 is y=(arcsin(x))/(sqrt(1-x^2))+(c_{1})/(sqrt(1-x^2))