You visited us 82 times! Enjoying our articles? Unlock Full Access!

Logarithmic Differentiation

The solution ...

Question

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

- \frac{1}{x y} = c

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

- \frac{1}{x y} +

logy

= c

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

\frac{1}{x y} + l o g y = c

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

l o g y = c x

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

Open in App

Solution

The correct option is A $- \frac{1}{x y} +$ logy $= c$
Given, $y d x + (x + x^{2} y) d y = 0$
$\Rightarrow y d x + x d y = - x^{2} y d y$
$\Rightarrow \frac{y d x + x d y}{x^{2} y^{2}} = - \frac{1}{y} d y$
$\Rightarrow d (- \frac{1}{x y}) = - \frac{1}{y} d y$
Integrating both sides, we get
$- \frac{1}{x y} = - log y + c$
$\Rightarrow - \frac{1}{x y} + log y = c$

Suggest Corrections

Similar questions

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)

(y + x + 5) d y = (y - x + 1) d x

(1 + e^{\frac{x}{y}}) d x + (1 - \frac{x}{y}) e^{\frac{x}{y}} d y = 0

c

x \frac{d y}{d x} = y (l o g y - l o g x + 1)

y = \frac{c - x}{1 + c x}

(1 + x^{2}) \frac{d y}{d x} + (1 + y^{2}) = 0

Join BYJU'S Learning Program