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

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

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

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

y = \frac{e ^{c_{1}} x + e ^{c_{1}}}{x - 1}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{c_{1}} x + e ^{c_{1}}}{x - 1}

y = \frac{e ^{c_{1}} x + e ^{c_{1}}}{x - 1}

\int_{0}^{3} \frac{1}{9 + x ^{2}} d x

x^{2} - tan (x)

\frac{d y}{d x}, x^{2} - y^{2} = 64

\frac{d}{d x} (sec (\frac{π}{4}))

cos (\frac{x + 1}{x - 1})

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