What is the solution for yy^'=sin(x) ?

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

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yy^'=sin(x)

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

y = - 2 cos (x) + c_{1}, y = - - 2 cos (x) + c_{1}

Solution steps

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

Solve separable ODE:

y = - 2 cos (x) + c_{1}, y = - - 2 cos (x) + c_{1}

y = - 2 cos (x) + c_{1}, y = - - 2 cos (x) + c_{1}

x \frac{d y}{d x} + y^{2} = 1

\frac{d x}{d t} = 60 - \frac{x}{100}, x = 2000, t = 0

x y^{2} cos (x) d x - 3 x^{2} y cos (x) d y = 0

\frac{d y}{d x} - y (6 x + 1) = 0

y^{^{'}} = x y - 8 x, y (0) = 9

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