What is the y^'(x)=(3x^2)/(3y^2-11),y(1)=0 ?

The y^'(x)=(3x^2)/(3y^2-11),y(1)=0 is -(3x^2)/(11)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'(x)=(3x^2)/(3y^2-11),y(1)=0

y^{^{'}} (x) = \frac{3 x ^{2}}{3 y ^{2} - 11}, y (1) = 0

- \frac{3 x ^{2}}{11}

Solution steps

\frac{3 x ^{2}}{3 y ^{2} - 11} = \frac{3 x ^{2}}{3 y ^{2} - 11}

Constant function results in a constant

y (1) = 0

= \frac{3 x ^{2}}{3 \cdot 0 ^{2} - 11}

Simplify

\frac{3 x ^{2}}{3 \cdot 0 ^{2} - 11} : - \frac{3 x ^{2}}{11}

= - \frac{3 x ^{2}}{11}

x^{2} - 3 x + 3

\frac{d y}{d x} = 2 x + 3 y - 1

\int (y^{3}) d y

\int \frac{7 ( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )} d x

y = 3 x^{2} + 4

What is the y^'(x)=(3x^2)/(3y^2-11),y(1)=0 ?
The y^'(x)=(3x^2)/(3y^2-11),y(1)=0 is -(3x^2)/(11)