Differential equation represents $2 x y d x + (y^{2} - x^{2}) d y = 0$

line

No worries! We‘ve got your back. Try BYJU‘S free classes today!

circle

No worries! We‘ve got your back. Try BYJU‘S free classes today!

hyperbola

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C None of these
Given, $2 x y d x + (y^{2} - x^{2}) d y = 0$
$\Rightarrow (y^{2} - x^{2}) \frac{d y}{d x} + 2 x y = 0$
Put $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
$∴ (- x^{2} + x^{2} v^{2}) (x \frac{d v}{d x} + v) + 2 x^{2} v = 0$

$\Rightarrow \frac{d v}{d x} = \frac{- v^{3} - v}{x (v^{2} - 1)} \Rightarrow \frac{(v^{2} - 1)}{(- v^{3} - v)} \frac{d v}{d x} = \frac{1}{x}$
Integrating both sides

$\int \frac{(v^{2} - 1)}{(- v^{3} - v)} \frac{d v}{d x} d x = \int \frac{1}{x} d x$
$\Rightarrow - log (v^{2} + 1) + log v = log x + c$
$\Rightarrow log (\frac{v}{1 + v^{2}}) = log x + log C \Rightarrow y = C (x^{2} + y^{2})$

Suggest Corrections

Similar questions

(1 + x^{2}) d y + 2 x y d x = cot x d x; x \neq 0

y (1 + x^{2}) d y + 2 x y d x = cot x d x

x^{2} d y = - 2 x y d x

(1 + x^{2}) d y + 2 x y d x = cot x d x

For the given differential equation find the general solution.

$(1 + x^{2}) d y + 2 x y d x = c o t x d x$

Join BYJU'S Learning Program