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

The solution for y^'=(tan(x))/y ,y(0)=-1 is y=-sqrt(-2ln(cos(x))+1)

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y^'=(tan(x))/y ,y(0)=-1

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

y = - - 2 ln (cos (x)) + 1

Solution steps

y^{'} (x) = \frac{tan ( x )}{y}

Solve separable ODE:

y = - - 2 ln (cos (x)) + 1

y = - - 2 ln (cos (x)) + 1

\frac{d y}{d x} = \frac{1 - x}{2 + y}

\frac{d y}{d x} = e^{2 x + 2 y}

\frac{d y}{d x} = \frac{2 y}{x - 1}

\frac{d y}{d x} = - \frac{2 x}{6 y}

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

What is the solution for y^'=(tan(x))/y ,y(0)=-1 ?
The solution for y^'=(tan(x))/y ,y(0)=-1 is y=-sqrt(-2ln(cos(x))+1)