What is the (dy)/(dx)=(1+x^2)/(1-y^2),y=0,x=0 ?

The (dy)/(dx)=(1+x^2)/(1-y^2),y=0,x=0 is 1/(1-y(0)^2)

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(dy)/(dx)=(1+x^2)/(1-y^2),y=0,x=0

\frac{d y}{d x} = \frac{1 + x ^{2}}{1 - y ^{2}}, y = 0, x = 0

\frac{1}{1 - y ( 0 ) ^{2}}

Solution steps

Find

\frac{1 + x ^{2}}{1 - y ^{2}}

given

y = 0, x = 0

Substitute

x = 0

y : 0

= \frac{1 + 0 ^{2}}{1 - y ( 0 ) ^{2}}

Expand

\frac{1 + 0 ^{2}}{1 - y ( 0 ) ^{2}} : \frac{1}{1 - y ( 0 ) ^{2}}

= \frac{1}{1 - y ( 0 ) ^{2}}

f^{^{'}} (x) + 2 x f (x) = 0, f (0) = 3

- 2 y \frac{d y}{d x} + x^{2} + 2 x^{4} = 0

\frac{d y}{d x} = \frac{- 0.8}{cos ( y )}, y (0) = \frac{π}{2}

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

y^{^{'}} sin (2 y) - cos (x) = 0

What is the (dy)/(dx)=(1+x^2)/(1-y^2),y=0,x=0 ?
The (dy)/(dx)=(1+x^2)/(1-y^2),y=0,x=0 is 1/(1-y(0)^2)