What is the solution for (1-x^2)dy=ydx ?

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

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(1-x^2)dy=ydx

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

y = e^{c_{1} + \frac{l n ( x + 1 ) - l n ( x - 1 )}{2}}

Solution steps

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

Solve separable ODE:

y = e^{c_{1} + \frac{l n ( x + 1 ) - l n ( x - 1 )}{2}}

y = e^{c_{1} + \frac{l n ( x + 1 ) - l n ( x - 1 )}{2}}

\frac{d y}{d x} = - \frac{0.2}{cos ( y )}, y (0) = \frac{π}{4}

y^{^{'}} + 6 y^{3} = 0

\frac{d y}{d x} = 3 x^{2} (y^{2} + 1), y (0) = 1

\frac{d v}{d t} = 0.075 v - 9000

2 x d x + 2 y d y = 0

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