What is the solution for y^'=(y*ln(x))/(x*ln(y)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(yln(x))/(xln(y))

y^{^{'}} = \frac{y \cdot ln ( x )}{x \cdot ln ( y )}

y = e^{l n^{2} (x) + c_{1}}, y = e^{- l n^{2} (x) + c_{1}}

Solution steps

y^{'} (x) = \frac{y \cdot ln ( x )}{x \cdot ln ( y )}

Solve separable ODE:

y = e^{l n^{2} (x) + c_{1}}, y = e^{- l n^{2} (x) + c_{1}}

y = e^{l n^{2} (x) + c_{1}}, y = e^{- l n^{2} (x) + c_{1}}

n = 1 \sum \infty \frac{x ^{n}}{9 ^{n}}

\frac{d y}{d t} + 2 y = 3 \cdot sin (6 t)

x \to - 2 - lim (\frac{- 1}{x ^{2} - 4})

x \to 7 lim (\frac{5}{x - 7})

\frac{d}{d x} ((y) x^{2})

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