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

The solution for xy^'+2y-2x^2y^2=0 is y= 1/(x^2(-2ln(x)+c_{1))}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

xy^'+2y-2x^2y^2=0

x y^{^{'}} + 2 y - 2 x^{2} y^{2} = 0

y = \frac{1}{x ^{2} ( - 2 ln ( x ) + c _{1} )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{1}{x ^{2} ( - 2 ln ( x ) + c _{1} )}

y = \frac{1}{x ^{2} ( - 2 ln ( x ) + c _{1} )}

f (x) = (2 x - 3)^{4} (x^{2} + x + 1)^{5}

\int \frac{1}{x ^{2} + 6 x + 13} d x

\int_{1}^{4} 3 e^{x} d x

\frac{d}{d x} (tan (π))

x = 1, y = 6 x (2 - x^{2})^{2}

What is the solution for xy^'+2y-2x^2y^2=0 ?
The solution for xy^'+2y-2x^2y^2=0 is y= 1/(x^2(-2ln(x)+c_{1))}