What is the solution for y^'=-(2x+ln(y))(y/x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=-(2x+ln(y))(y/x)

y^{^{'}} = - (2 x + ln (y)) (\frac{y}{x})

y = e^{\frac{c _{1} - x ^{2}}{x}}

Solution steps

y^{'} (x) = - (2 x + ln (y)) (\frac{y}{x})

Solve using substitution:

y = e^{\frac{c _{1} - x ^{2}}{x}}

y = e^{\frac{c _{1} - x ^{2}}{x}}

\frac{x ^{3}}{3} - x^{2} - 15 x

x \to 0 lim (2 - 6 x)

\frac{d}{d x} (2 e^{x} + 2 x e^{x})

\frac{1}{s} e^{- s}

\int 5 x (2 x + 5) (x - 3) d x

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