You visited us 2 times! Enjoying our articles? Unlock Full Access!

Homogeneous Function

The solution ...

Question

The solution of differential equation $y d x + (x - y^{2}) d y = 0$

e^{\frac{y}{x}} = sin x + c

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

y = c x log x

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

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

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

cos (\frac{y - 2}{x}) = a

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

Open in App

Solution

The correct option is C $x = \frac{y^{2}}{3} + \frac{c}{y}$
$y d x + (x - y^{2}) d y = 0$
$y d x + x d y - y^{2} d y = 0$
$d (x y) - y^{2} d y = 0$
Integrating, we have
$x y - \frac{y^{3}}{3} = c$

This can also be written as
$x = \frac{y^{3}}{3} + \frac{c}{y}$

Suggest Corrections

Similar questions

y d x - x d y = y^{2} tan (\frac{x}{y}) d x

C

y e^{\frac{x}{y}} d x = (x e^{\frac{x}{y}} + y^{2}) d y, (y \neq 0)

^{'} C^{'}

y d x - (x + y) d y = 0

y = e^{x} + 1

y^{''} - y = 0

y = x^{2} + 2 x + C

y^{'} - 2 x - 2 = 0

y = cos x + C

y^{'} + sin x = 0

y = \sqrt{1 + x^{2}}

y^{'} = \frac{x y}{1 + x^{2}}

y = A x

x y = y (x \neq 0)

y = x sin x

x y = y + x \sqrt{x^{2} y^{2}} (x \neq 0 a n d x > y o r x < y)

x y = log y + C

y^{'} = \frac{y^{2}}{1 - x y} (x y \neq 1)

y - cos y = x

(y sin y + cos y + x) y^{'} = y

x + y = {tan}^{- 1} y

y^{2} y^{'} + y^{2} + 1 = 0

y = \sqrt{a^{2} - x^{2}} x ϵ (- a, a)

x + y \frac{d y}{d x} = 0 (y \neq 0)

(x^{2} s i n^{3} y - y^{2} c o s x) d x + (x^{3} c o s y s i n^{2} y - 2 y s i n x) d y = 0

Join BYJU'S Learning Program