Solve the differential equation:
$(x y^{2} + x) d x + (x^{2} y + y) d y = 0$

Open in App

Solution

$(x y^{2} + x) d x + (x^{2} y + y) d y = 0$
$x (y^{2} + 1) d x + y (x^{2} + 1) d y = 0$
$\Rightarrow x (y^{2} + 1) d x = - y (x^{2} + 1) d x$
$\Rightarrow \frac{x}{x^{2} + 1} d x = \frac{- y}{y^{2} + 1} d y$
$\Rightarrow \frac{2 x}{x^{2} + 1} d x = \frac{- 2 y}{y^{2} + 1} d y$
$\Rightarrow l o g | x^{2} + 1 | = - l o g | y^{2} + 1 | + c$
$\Rightarrow l o g | x^{2} + 1 | + l o g | y^{2} + 1 | = c$
$\Rightarrow | x^{2} + 1 | | y^{2} + 1 | = e^{c}$
$\Rightarrow (x^{2} + 1) (y^{2} + 1) = \pm e^{c}$

Suggest Corrections

Similar questions

x d x + y d y = x d y - y d x

cosec x \log y \frac{d y}{d x} + x^{2} y^{2} = 0

The solution of the differential equation x dx + y dy = 0 is:

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

\frac{\sqrt{x} d x + \sqrt{y} d y}{\sqrt{x} d x - \sqrt{y} d y} = \sqrt{\frac{y^{3}}{x^{3}}}

Join BYJU'S Learning Program