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

The y^'(x)=(ycos(x))/(1+3y^3),y(0)=1 is (cos(x))/4

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y^'(x)=(ycos(x))/(1+3y^3),y(0)=1

y^{^{'}} (x) = \frac{y cos ( x )}{1 + 3 y ^{3}}, y (0) = 1

\frac{cos ( x )}{4}

Solution steps

\frac{y cos ( x )}{1 + 3 y ^{3}} = \frac{y cos ( x )}{1 + 3 y ^{3}}

Constant function results in a constant

y (0) = 1

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

Expand

\frac{1 \cdot cos ( x )}{1 + 3 \cdot 1 ^{3}} : \frac{cos ( x )}{4}

= \frac{cos ( x )}{4}

What is the y^'(x)=(ycos(x))/(1+3y^3),y(0)=1 ?
The y^'(x)=(ycos(x))/(1+3y^3),y(0)=1 is (cos(x))/4