What is the solution for (4+y^2)dx+(9+x^2)dy=0 ?

The solution for (4+y^2)dx+(9+x^2)dy=0 is y=2tan(-2/3 arctan(x/3)-c_{1})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(4+y^2)dx+(9+x^2)dy=0

(4 + y^{2}) d x + (9 + x^{2}) d y = 0

y = 2 tan (- \frac{2}{3} arctan (\frac{x}{3}) - c_{1})

Solution steps

(4 + y^{2}) d x + (9 + x^{2}) d y = 0

Solve separable ODE:

y = 2 tan (- \frac{2}{3} arctan (\frac{x}{3}) - c_{1})

y = 2 tan (- \frac{2}{3} arctan (\frac{x}{3}) - c_{1})

\int \frac{x ^{4} + 2 x + 6}{x ^{3} + x ^{2} - 2 x} d x

\int \frac{11}{11 + e ^{x}} d x

\frac{d}{d t} (tan (e^{2 t}) + e^{t a n (2 t)})

x \to 0 - lim (\frac{1}{x} - \frac{1}{∣ x ∣})

\frac{d}{d x} (x^{3} - 2 x^{2} + x + 1)

What is the solution for (4+y^2)dx+(9+x^2)dy=0 ?
The solution for (4+y^2)dx+(9+x^2)dy=0 is y=2tan(-2/3 arctan(x/3)-c_{1})