What is the solution for x(dy)/(dx)=y(ln(y)-ln(x)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x(dy)/(dx)=y(ln(y)-ln(x))

x \frac{d y}{d x} = y (ln (y) - ln (x))

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

Solution steps

x \frac{d y}{d x} = y (ln (y) - ln (x))

Solve using substitution:

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

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

\frac{d}{d x} (\frac{ln ( x ) - 1}{ln ^{2} ( x )})

\frac{12}{1 + x ^{4}}, 6 x^{2}

2 x^{x}

n = 0 \sum \infty \frac{2 ^{2}}{3 ^{2}}

\frac{\partial}{\partial s} (- 3 s^{2} t^{2})

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