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

The solution for y^'=((y^2-1))/(2x),y(1)=-2 is y=-1/(3x-1)-(3x)/(3x-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

x \to 2 lim (x)

x, x^{3}, 4 x, 0

\frac{1}{6 s + 1}

f (x) = \frac{1}{x}, g (x) = - \frac{1}{3} x + 3, [0, 12]

What is the solution for y^'=((y^2-1))/(2x),y(1)=-2 ?
The solution for y^'=((y^2-1))/(2x),y(1)=-2 is y=-1/(3x-1)-(3x)/(3x-1)