What is the solution for e^{(2x)}y^2y^'=1 ?

The solution for e^{(2x)}y^2y^'=1 is y=\sqrt[3]{(-3e^{-2x}+c_{1})/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

e^{(2x)}y^2y^'=1

e^{(2 x)} y^{2} y^{^{'}} = 1

Solution steps

e^{(2 x)} y^{2} y^{'} (x) = 1

Solve separable ODE:

\frac{d y}{d x} + \frac{3 y}{x} = \frac{3}{x}

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

x sin (x) y^{^{'}} - (sin (x) + x cos (x)) y = 0

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

\frac{d y}{d x} = \frac{x}{y} - \frac{x}{1 + y}, y (0) = 1

What is the solution for e^{(2x)}y^2y^'=1 ?
The solution for e^{(2x)}y^2y^'=1 is y=\sqrt[3]{(-3e^{-2x}+c_{1})/2}