What is the solution for (17y^2-xy)dx+x^2dy=0 ?

The solution for (17y^2-xy)dx+x^2dy=0 is y= x/(17ln(x)+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(17y^2-xy)dx+x^2dy=0

(17 y^{2} - x y) d x + x^{2} d y = 0

y = \frac{x}{17 ln ( x ) + c _{1}}

Solution steps

(17 y^{2} - x y) d x + x^{2} d y = 0

Solve Bernoulli ODE:

y = \frac{x}{17 ln ( x ) + c _{1}}

y = \frac{x}{17 ln ( x ) + c _{1}}

\int \frac{e ^{3 x}}{5 + e ^{3 x}} d x

\frac{d}{d x} (arccsc (- 6 x^{2}))

x \to 0 lim (\frac{x}{x ^{2} + 3 x})

\int u^{5} d u

f (x) = \frac{x ( 2 - ln ( 2 ) x )}{2 ^{x}}

What is the solution for (17y^2-xy)dx+x^2dy=0 ?
The solution for (17y^2-xy)dx+x^2dy=0 is y= x/(17ln(x)+c_{1)}