What is the y^'(x)=(ycos(x))/(1+2y^2),y(0)=1 ?

The y^'(x)=(ycos(x))/(1+2y^2),y(0)=1 is (cos(x))/3

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y^'(x)=(ycos(x))/(1+2y^2),y(0)=1

y^{^{'}} (x) = \frac{y cos ( x )}{1 + 2 y ^{2}}, y (0) = 1

\frac{cos ( x )}{3}

Solution steps

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

Constant function results in a constant

y (0) = 1

= \frac{1 \cdot cos ( x )}{1 + 2 \cdot 1 ^{2}}

Expand

\frac{1 \cdot cos ( x )}{1 + 2 \cdot 1 ^{2}} : \frac{cos ( x )}{3}

= \frac{cos ( x )}{3}

\frac{d t}{d y} = 2 t \cdot y^{2}

7 x y (\frac{d y}{d x}) = 4

\frac{d x}{d t} = \frac{2}{x}, x (0) = 8

y^{^{'}} sec (t) = \frac{- y ^{21}}{20}

sin (y) \cdot cos (x) \cdot y^{^{'}} = - cos (y) \cdot sin (x)

What is the y^'(x)=(ycos(x))/(1+2y^2),y(0)=1 ?
The y^'(x)=(ycos(x))/(1+2y^2),y(0)=1 is (cos(x))/3