What is the solution for y^'=(ln(x))/(xy),y(1)=2 ?

The solution for y^'=(ln(x))/(xy),y(1)=2 is y=sqrt(ln^2(x)+4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(ln(x))/(xy),y(1)=2

y^{^{'}} = \frac{ln ( x )}{x y}, y (1) = 2

y = ln^{2} (x) + 4

Solution steps

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

Solve separable ODE:

y = ln^{2} (x) + 4

y = ln^{2} (x) + 4

- (\frac{d x}{d y}) = \frac{( x ^{2} y ^{2} )}{( 1 + x )}

\frac{d y}{d t} = e^{- y} (2 t - 4), y (2) = 0

\frac{d h}{d t} = 1 - 0.5 h

\frac{( x ^{2} - 9 )}{x} \frac{d y}{d x} + y = 0, y (5) = 1

\frac{d y}{d x} = 7 (5 - y)

What is the solution for y^'=(ln(x))/(xy),y(1)=2 ?
The solution for y^'=(ln(x))/(xy),y(1)=2 is y=sqrt(ln^2(x)+4)