What is the solution for y^'=(xy^2-0.5sin(2x))/(y(1-x^2)),y(0)=2 ?

The solution for y^'=(xy^2-0.5sin(2x))/(y(1-x^2)),y(0)=2 is y=sqrt((cos(2x)+7)/(2(-x^2+1)))

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y^'=(xy^2-0.5sin(2x))/(y(1-x^2)),y(0)=2

y^{^{'}} = \frac{x y ^{2} - 0.5 sin ( 2 x )}{y ( 1 - x ^{2} )}, y (0) = 2

y = \frac{cos ( 2 x ) + 7}{2 ( - x ^{2} + 1 )}

Solution steps

y^{'} (x) = \frac{x y ^{2} - 0.5 sin ( 2 x )}{y ( 1 - x ^{2} )}

Solve Bernoulli ODE:

y = \frac{cos ( 2 x ) + 7}{2 ( - x ^{2} + 1 )}

y = \frac{cos ( 2 x ) + 7}{2 ( - x ^{2} + 1 )}

y^{^{'}} = (t + y)^{2} - 1, y (2) = 6

\frac{d}{d x} ((\frac{x + 2}{x - 2})^{5})

\frac{d}{d x} (2 x^{2} + y^{2})

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

\frac{d y}{d x} = 3 x y

What is the solution for y^'=(xy^2-0.5sin(2x))/(y(1-x^2)),y(0)=2 ?
The solution for y^'=(xy^2-0.5sin(2x))/(y(1-x^2)),y(0)=2 is y=sqrt((cos(2x)+7)/(2(-x^2+1)))