What is the solution for y^'sec(x)=y^2,y((13pi)/2)= 1/3 ?

The solution for y^'sec(x)=y^2,y((13pi)/2)= 1/3 is y=-1/(sin(x)-4)

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y^'sec(x)=y^2,y((13pi)/2)= 1/3

y^{^{'}} sec (x) = y^{2}, y (\frac{13 π}{2}) = \frac{1}{3}

y = - \frac{1}{sin ( x ) - 4}

Solution steps

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

Solve separable ODE:

y = - \frac{1}{sin ( x ) - 4}

y = - \frac{1}{sin ( x ) - 4}

\frac{d y}{d x} = - 16 x y

(\frac{y}{x + 6 x}) + (ln (x) - 2) y^{^{'}} = 0

\frac{d y}{d x} = e^{3 x - 5 y}

\frac{d y}{d x} = \frac{( 1 + 4 y ^{2} )}{e ^{x}}

\frac{d y}{d x} = - 8 x y^{2}, y (0) = 1

What is the solution for y^'sec(x)=y^2,y((13pi)/2)= 1/3 ?
The solution for y^'sec(x)=y^2,y((13pi)/2)= 1/3 is y=-1/(sin(x)-4)