What is the solution for y^'= 1/(xy) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'= 1/(xy)

y^{^{'}} = \frac{1}{x y}

y = 2 ln (x) + c_{1}, y = - 2 ln (x) + c_{1}

Solution steps

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

Solve separable ODE:

y = 2 ln (x) + c_{1}, y = - 2 ln (x) + c_{1}

y = 2 ln (x) + c_{1}, y = - 2 ln (x) + c_{1}

f (x) = 8 x^{2} - 3, at x = 3

f (0) = \frac{cos ( x )}{x ^{2} - 1}

\int_{1}^{4} 7 e^{x} d x

n = 1 \sum \infty \frac{2 ^{n}}{1 0 ^{n}}

\frac{\partial ^{2}}{\partial y ^{2}} (x y^{10} + x^{2} + y^{4})

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