What is the solution for y^'=(sin(x))/y ,y(pi/2)=1 ?

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

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y^'=(sin(x))/y ,y(pi/2)=1

y^{^{'}} = \frac{sin ( x )}{y}, y (\frac{π}{2}) = 1

y = - 2 cos (x) + 1

Solution steps

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

Solve separable ODE:

y = - 2 cos (x) + 1

y = - 2 cos (x) + 1

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

y^{^{'}} (x) = \frac{4 + 3 x ^{2}}{3 y ^{2} - 12 y}, y (0) = 2

(y cos (x)) d x + (3 sin (x)) d y = 0

y^{5} \frac{d y}{d x} = 2 x^{4} y^{6} - 4 x^{4}

\frac{d y}{d x} = 4 x^{4} y

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