What is the solution for y^'sec(x)=8y^2,y(pi/2)= 1/3 ?

The solution for y^'sec(x)=8y^2,y(pi/2)= 1/3 is y=-1/(8sin(x)-11)

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y^'sec(x)=8y^2,y(pi/2)= 1/3

y^{^{'}} sec (x) = 8 y^{2}, y (\frac{π}{2}) = \frac{1}{3}

y = - \frac{1}{8 sin ( x ) - 11}

Solution steps

y^{'} (x) sec (x) = 8 y^{2}

Solve separable ODE:

y = - \frac{1}{8 sin ( x ) - 11}

y = - \frac{1}{8 sin ( x ) - 11}

\frac{d P}{d t} = 7 Pt, P (1) = 7

\frac{d y}{d t} = cos^{2} (y)

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

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

What is the solution for y^'sec(x)=8y^2,y(pi/2)= 1/3 ?
The solution for y^'sec(x)=8y^2,y(pi/2)= 1/3 is y=-1/(8sin(x)-11)