What is the solution for 0=2x+1-2yy^',y(-2)=-1 ?

The solution for 0=2x+1-2yy^',y(-2)=-1 is y=-sqrt(x^2+x-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0=2x+1-2yy^',y(-2)=-1

0 = 2 x + 1 - 2 y y^{^{'}}, y (- 2) = - 1

y = - x^{2} + x - 1

Solution steps

0 = 2 x + 1 - 2 y y^{'} (x)

Solve separable ODE:

y = - x^{2} + x - 1

y = - x^{2} + x - 1

x \to 1 lim (arctan (\frac{x}{2}))

\frac{d}{d x} (sin (2 \cdot x))

\frac{d}{d x} (5 + cos (x^{2}))

\int e^{5 y} d y

\int 12 gx d x

What is the solution for 0=2x+1-2yy^',y(-2)=-1 ?
The solution for 0=2x+1-2yy^',y(-2)=-1 is y=-sqrt(x^2+x-1)