What is the solution for y^'=(x^2+2)/y ?

The solution for y^'=(x^2+2)/y is y=sqrt((2x^3)/3+4x+c_{1)},y=-sqrt((2x^3)/3+4x+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'=(x^2+2)/y

y^{^{'}} = \frac{x ^{2} + 2}{y}

y = \frac{2 x ^{3}}{3} + 4 x + c_{1}, y = - \frac{2 x ^{3}}{3} + 4 x + c_{1}

Solution steps

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

Solve separable ODE:

y = \frac{2 x ^{3}}{3} + 4 x + c_{1}, y = - \frac{2 x ^{3}}{3} + 4 x + c_{1}

y = \frac{2 x ^{3}}{3} + 4 x + c_{1}, y = - \frac{2 x ^{3}}{3} + 4 x + c_{1}

What is the solution for y^'=(x^2+2)/y ?
The solution for y^'=(x^2+2)/y is y=sqrt((2x^3)/3+4x+c_{1)},y=-sqrt((2x^3)/3+4x+c_{1)}