What is the solution for y^'=(2x)/(1+2y),y(2)=-0.2 ?

The solution for y^'=(2x)/(1+2y),y(2)=-0.2 is y=(-5+sqrt(100x^2-391))/(10)

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)=-0.2

y^{^{'}} = \frac{2 x}{1 + 2 y}, y (2) = - 0.2

y = \frac{- 5 + 100 x ^{2} - 391}{10}

Solution steps

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

Solve separable ODE:

y = \frac{- 5 + 100 x ^{2} - 391}{10}

y = \frac{- 5 + 100 x ^{2} - 391}{10}

\frac{d y}{d x} = 9 x y^{2}

y^{^{'}} = \frac{2 x}{1 + 2 y}, y (2) = - 0.5

\frac{d y}{d x} = 7 x y - 7 x

(3 x y^{2} - 6 y) d x + (3 x^{2} y - 6 x) d y = 0

d y = (x - 1) d x + y (x - 1) d x

What is the solution for y^'=(2x)/(1+2y),y(2)=-0.2 ?
The solution for y^'=(2x)/(1+2y),y(2)=-0.2 is y=(-5+sqrt(100x^2-391))/(10)