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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = 3 6 (\frac{1}{3} (1 + x)^{\frac{3}{2}} - 1 + x) + c_{1}

Solution steps

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

Solve separable ODE:

y = 3 6 (\frac{1}{3} (1 + x)^{\frac{3}{2}} - 1 + x) + c_{1}

y = 3 6 (\frac{1}{3} (1 + x)^{\frac{3}{2}} - 1 + x) + c_{1}

\frac{d y}{d x} = \frac{6 y}{x}, x > 0

y^{^{'}} = 3 \frac{y}{x - 1}

y^{^{'}} + 6 y = 4, y (0) = 0

x \frac{d y}{d x} = y x - x y cos (x), y = 1, x = 0

y^{^{'}} = y \cdot e^{- x}

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