What is the y^'(x)=(x^2)/(1-y^2),y(4)=2 ?

The y^'(x)=(x^2)/(1-y^2),y(4)=2 is -(x^2)/3

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y^'(x)=(x^2)/(1-y^2),y(4)=2

y^{^{'}} (x) = \frac{x ^{2}}{1 - y ^{2}}, y (4) = 2

- \frac{x ^{2}}{3}

Solution steps

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

Constant function results in a constant

y (4) = 2

= \frac{x ^{2}}{1 - 2 ^{2}}

Simplify

\frac{x ^{2}}{1 - 2 ^{2}} : - \frac{x ^{2}}{3}

= - \frac{x ^{2}}{3}

\int 2 x e^{x} d x

y = (x^{2} + 1)^{3}

\int \frac{1 + cos ( 2 x )}{sin ^{2} ( 2 x )} d x

\frac{\partial}{\partial y} (x + 3 y)

(- 1.3) (3.5)

What is the y^'(x)=(x^2)/(1-y^2),y(4)=2 ?
The y^'(x)=(x^2)/(1-y^2),y(4)=2 is -(x^2)/3