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

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

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y^'=xy^2sin(x),y(0)=2

y^{^{'}} = x y^{2} sin (x), y (0) = 2

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

Solution steps

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

Solve separable ODE:

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

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

x^{2} y^{^{'}} - 2 y = 0

f^{^{'}} (x) = - 2 \cdot f (x)

\frac{d y}{d x} = 7 y - 1

\frac{d y}{d x} = \frac{cos ( 3 x )}{y ^{2}}

(y^{2} sec (2 x) tan (2 x)) d x + y (sec (2 x) + 2) d y = 0

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