What is the solution for y^2e^{2x}dx-4ydy-4ye^{2x}dy=0 ?

The solution for y^2e^{2x}dx-4ydy-4ye^{2x}dy=0 is y=c_{1}\sqrt[8]{e^{2x}+1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^2e^{2x}dx-4ydy-4ye^{2x}dy=0

y^{2} e^{2 x} d x - 4 y d y - 4 y e^{2 x} d y = 0

y = c_{1} 8 e^{2 x} + 1

Solution steps

y^{2} e^{2 x} d x - 4 y d y - 4 y e^{2 x} d y = 0

Solve separable ODE:

y = c_{1} 8 e^{2 x} + 1

y = c_{1} 8 e^{2 x} + 1

y^{3} - y^{^{'}} + 0.5 y = 0, y (0) = \frac{1}{3}

d v = \frac{x}{4 + x ^{2}} d x

x^{^{'}} = (1 - x) (1 - cos (π) x)

y^{^{'}} = - \frac{0.0491}{c} \cdot 19.62 (y + 1)

(y)^{6} (\frac{\frac{d y}{d x} - 5}{3}) = tan (3 x) - \frac{5}{3} (y)^{6}

What is the solution for y^2e^{2x}dx-4ydy-4ye^{2x}dy=0 ?
The solution for y^2e^{2x}dx-4ydy-4ye^{2x}dy=0 is y=c_{1}\sqrt[8]{e^{2x}+1}