What is the y^'(x)=(1-y^2)tan(x),y(0)=sqrt(3) ?

The y^'(x)=(1-y^2)tan(x),y(0)=sqrt(3) is tan(x)-tan(x)y^2

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y^'(x)=(1-y^2)tan(x),y(0)=sqrt(3)

y^{^{'}} (x) = (1 - y^{2}) tan (x), y (0) = 3

tan (x) - tan (x) y^{2}

Solution steps

(1 - y^{2}) tan (x) = (1 - y^{2}) tan (x)

Constant function results in a constant

y (0) = 3

= (3)^{2}

Expand

(1 - y^{2}) tan (x) : tan (x) - tan (x) y^{2}

= tan (x) - tan (x) y^{2}

\frac{d y}{d x} = \frac{- 0.9}{cos ( y )}, y (0) = \frac{π}{3}

e^{2 x} d y + 2 y d x = 0

\frac{d v}{d t} = 2 - 8 v

\frac{d y}{d x} = \frac{e ^{x}}{cos ( y )}

7 x y (\frac{d y}{d x}) = 3

What is the y^'(x)=(1-y^2)tan(x),y(0)=sqrt(3) ?
The y^'(x)=(1-y^2)tan(x),y(0)=sqrt(3) is tan(x)-tan(x)y^2