What is the solution for y^'sec(x)=5y^2,y(pi/2)= 1/2 ?

The solution for y^'sec(x)=5y^2,y(pi/2)= 1/2 is y=-1/(5sin(x)-7)

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y^'sec(x)=5y^2,y(pi/2)= 1/2

y^{^{'}} sec (x) = 5 y^{2}, y (\frac{π}{2}) = \frac{1}{2}

y = - \frac{1}{5 sin ( x ) - 7}

Solution steps

y^{'} (x) sec (x) = 5 y^{2}

Solve separable ODE:

y = - \frac{1}{5 sin ( x ) - 7}

y = - \frac{1}{5 sin ( x ) - 7}

\frac{d Q}{d t} + \frac{Q}{20} = 0

\frac{d y}{d x} x y + 2 x + 3 x^{2} = 4

y^{2} \frac{d y}{d x} + 2 x y^{3} = 6 x, y (0) = 1

\frac{d y}{d x} = 3.2 y^{\frac{2}{3}}, x (0) = 1, y (0) = 13

\frac{d}{d x} ((y) x) + 2 y = 0

What is the solution for y^'sec(x)=5y^2,y(pi/2)= 1/2 ?
The solution for y^'sec(x)=5y^2,y(pi/2)= 1/2 is y=-1/(5sin(x)-7)