What is the solution for y^'=(x(x^2+1))/((4y^2)) ?

The solution for y^'=(x(x^2+1))/((4y^2)) is y=(\sqrt[3]{3x^4+6x^2+c_{1}})/(2\sqrt[3]{2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(x(x^2+1))/((4y^2))

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

y = \frac{3 3 x ^{4} + 6 x ^{2} + c _{1}}{2 3 2}

Solution steps

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

Solve separable ODE:

y = \frac{3 3 x ^{4} + 6 x ^{2} + c _{1}}{2 3 2}

y = \frac{3 3 x ^{4} + 6 x ^{2} + c _{1}}{2 3 2}

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

\frac{d x}{d t} = \frac{100 - x}{10}

(\frac{df}{d y}) f (y) = e^{- y}

4 y^{3} \frac{d y}{d x} - 5 x^{2} + 3 x = 0

y ln (x) \frac{d x}{d y} = (\frac{y + 1}{x})

What is the solution for y^'=(x(x^2+1))/((4y^2)) ?
The solution for y^'=(x(x^2+1))/((4y^2)) is y=(\sqrt[3]{3x^4+6x^2+c_{1}})/(2\sqrt[3]{2)}