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

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

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y^'=(sin(x))/y

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

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

Solution steps

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

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}

\int x sin (- 4 x) d x

\int \frac{1}{( x ^{2} + 25 ) ^{\frac{3}{2}}} d x

(4 x - y) d x + (6 y - x) d y = 0

\int (5 x^{2} - 6)^{\frac{3}{2}} d x

\frac{d}{d x} (\frac{12}{ln ( x )})

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