What is the solution for (2xy+3)dx+(x^2-1)dy=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2xy+3)dx+(x^2-1)dy=0

(2 x y + 3) d x + (x^{2} - 1) d y = 0

y = - \frac{3 x}{x ^{2} - 1} + \frac{c _{1}}{x ^{2} - 1}

Solution steps

(2 x y + 3) d x + (x^{2} - 1) d y = 0

Solve linear ODE:

y = - \frac{3 x}{x ^{2} - 1} + \frac{c _{1}}{x ^{2} - 1}

y = - \frac{3 x}{x ^{2} - 1} + \frac{c _{1}}{x ^{2} - 1}

3 x y^{^{'}} + 7 y = 0

\frac{e ^{10 x} + e ^{- 10 x}}{x}

\frac{d}{d x} (2 (ln (x)))

\int_{0}^{1.5} (\frac{3}{2} - x)^{2} d x

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 8} e^{4 t}

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