What is the solution for y^'sec(x)=5y^2,y((13pi)/2)= 1/8 ?

The solution for y^'sec(x)=5y^2,y((13pi)/2)= 1/8 is y=-1/(5sin(x)-13)

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y^'sec(x)=5y^2,y((13pi)/2)= 1/8

y^{^{'}} sec (x) = 5 y^{2}, y (\frac{13 π}{2}) = \frac{1}{8}

y = - \frac{1}{5 sin ( x ) - 13}

Solution steps

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

Solve separable ODE:

y = - \frac{1}{5 sin ( x ) - 13}

y = - \frac{1}{5 sin ( x ) - 13}

\frac{y ^{^{'}}}{y ^{\frac{1}{2}}} = - 10

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

(9 + x) y^{^{'}} = 2 y

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

\frac{d y}{d t} = \frac{t}{y}

What is the solution for y^'sec(x)=5y^2,y((13pi)/2)= 1/8 ?
The solution for y^'sec(x)=5y^2,y((13pi)/2)= 1/8 is y=-1/(5sin(x)-13)