What is the solution for 2y*y^'-y^2x-2x=0 ?

The solution for 2y*y^'-y^2x-2x=0 is y=sqrt(e^{(x^2)/2+c_{1)}-2},y=-sqrt(e^{(x^2)/2+c_{1)}-2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y*y^'-y^2x-2x=0

2 y \cdot y^{^{'}} - y^{2} x - 2 x = 0

y = e^{\frac{x ^{2}}{2} + c_{1}} - 2, y = - e^{\frac{x ^{2}}{2} + c_{1}} - 2

Solution steps

2 y \cdot y^{'} (x) - y^{2} x - 2 x = 0

Solve separable ODE:

y = e^{\frac{x ^{2}}{2} + c_{1}} - 2, y = - e^{\frac{x ^{2}}{2} + c_{1}} - 2

y = e^{\frac{x ^{2}}{2} + c_{1}} - 2, y = - e^{\frac{x ^{2}}{2} + c_{1}} - 2

y^{^{'}} = - \frac{2 y}{2 x + 100}

\frac{( x ^{2} - 9 )}{x} \cdot \frac{d y}{d x} + y = 0

\frac{d y}{d x} = \frac{x ^{2} + 7 x + 3}{y ^{2}}

(x y^{2} - x) d x + (2 x^{2} + 4) d y = 0

\frac{d y}{d x} = \frac{3 x ^{2} - e ^{x}}{2 y - 5}

What is the solution for 2y*y^'-y^2x-2x=0 ?
The solution for 2y*y^'-y^2x-2x=0 is y=sqrt(e^{(x^2)/2+c_{1)}-2},y=-sqrt(e^{(x^2)/2+c_{1)}-2}