What is the solution for (dy)/(dx)-tan(x)y=y^3,y(0)=1 ?

The solution for (dy)/(dx)-tan(x)y=y^3,y(0)=1 is y=(sec(x))/(sqrt(-2tan(x)+1))

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(dy)/(dx)-tan(x)y=y^3,y(0)=1

\frac{d y}{d x} - tan (x) y = y^{3}, y (0) = 1

y = \frac{sec ( x )}{- 2 tan ( x ) + 1}

Solution steps

\frac{d y}{d x} - tan (x) y = y^{3}

Solve Bernoulli ODE:

y = \frac{sec ( x )}{- 2 tan ( x ) + 1}

y = \frac{sec ( x )}{- 2 tan ( x ) + 1}

\int 5 x^{4} e^{x^{5}} d x

\frac{d}{d t} (t^{3} + t)

\frac{d}{d x} (x^{2} \cdot ln (x^{2}))

f (x) = \frac{1}{x + 6}, at x = 3

x^{3}, 2 x - 15.8 x^{2}

What is the solution for (dy)/(dx)-tan(x)y=y^3,y(0)=1 ?
The solution for (dy)/(dx)-tan(x)y=y^3,y(0)=1 is y=(sec(x))/(sqrt(-2tan(x)+1))