What is the solution for (dy)/(dx)=(2x+1)/(2y),y(-2)=-1 ?

The solution for (dy)/(dx)=(2x+1)/(2y),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 >

(dy)/(dx)=(2x+1)/(2y),y(-2)=-1

\frac{d y}{d x} = \frac{2 x + 1}{2 y}, y (- 2) = - 1

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

Solution steps

\frac{d y}{d x} = \frac{2 x + 1}{2 y}

Solve separable ODE:

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

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

2 y = 3 - x y^{^{'}}

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

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (10)

(3 x cos (3 x) + sin (3 x) - 3) d x + (2 y + 5) d y = 0

sin (2 x) d y - e^{y} d x = 0

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