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

The solution for yy^'=(1-y^2)sin(x) is y=sqrt(-e^{2cos(x)-c_{1)}+1},y=-sqrt(-e^{2cos(x)-c_{1)}+1}

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

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

y = - e^{2 c o s (x) - c_{1}} + 1, y = - - e^{2 c o s (x) - c_{1}} + 1

Solution steps

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

Solve separable ODE:

y = - e^{2 c o s (x) - c_{1}} + 1, y = - - e^{2 c o s (x) - c_{1}} + 1

y = - e^{2 c o s (x) - c_{1}} + 1, y = - - e^{2 c o s (x) - c_{1}} + 1

What is the solution for yy^'=(1-y^2)sin(x) ?
The solution for yy^'=(1-y^2)sin(x) is y=sqrt(-e^{2cos(x)-c_{1)}+1},y=-sqrt(-e^{2cos(x)-c_{1)}+1}