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

The solution for y^'=(2x)/(1+2y),y(2)=1 is y=(-1+sqrt(4x^2-7))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = \frac{- 1 + 4 x ^{2} - 7}{2}

Solution steps

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

Solve separable ODE:

y = \frac{- 1 + 4 x ^{2} - 7}{2}

y = \frac{- 1 + 4 x ^{2} - 7}{2}

\int_{1}^{\infty} e^{- s t} d t

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 4 )}

x (\frac{d y}{d x}) = \frac{( 1 - 4 y ^{2} )}{3 y}

x x + x^{2} x

\frac{\partial}{\partial x} (\frac{5}{x})

What is the solution for y^'=(2x)/(1+2y),y(2)=1 ?
The solution for y^'=(2x)/(1+2y),y(2)=1 is y=(-1+sqrt(4x^2-7))/2