What is the solution for (10y^2-xy)dx+(x^2)dy=0 ?

The solution for (10y^2-xy)dx+(x^2)dy=0 is y= x/(10ln(x)+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(10y^2-xy)dx+(x^2)dy=0

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

y = \frac{x}{10 ln ( x ) + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x}{10 ln ( x ) + c _{1}}

y = \frac{x}{10 ln ( x ) + c _{1}}

\int 2 x (x^{2} + 3)^{2} d x

24 y^{^{''}} + 12 y^{^{'}} = 0, y (3) = - 3, y^{^{'}} (- 4) = 4

(2 x + 9)^{2}

\int \frac{tan ( x )}{1 - sin ^{2} ( x )} d x

f (x) = lo g_{e} (10 x)

What is the solution for (10y^2-xy)dx+(x^2)dy=0 ?
The solution for (10y^2-xy)dx+(x^2)dy=0 is y= x/(10ln(x)+c_{1)}