Find the general solution of the differential equation: $(x^{2} - 1) \frac{d y}{d x} + 2 x y = 1$

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

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

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

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

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

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

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

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

Open in App

Solution

The correct option is A $y = \frac{x}{x^{2} - 1} + \frac{c}{x^{2} - 1}$
The differential equation can be written as $\frac{d y}{d x} + \frac{2 x}{x^{2} - 1} y = \frac{1}{x^{2} - 1}$
It is in the form of linear equation
The integrating factor will be $e^{\int \frac{2 x}{x^{2} - 1} d x} = e^{ln (x^{2} - 1)} = x^{2} - 1$
The general solution is $y (x^{2} - 1) = \int \frac{1}{x^{2} - 1} (x^{2} - 1) d x + c$ $($ where $c$ is constant $)$
$\Rightarrow y = \frac{x}{x^{2} - 1} + \frac{c}{x^{2} - 1}$

Suggest Corrections

Similar questions

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{(1 + x^{2})^{2}}$ is
(a) $y (1 + x^{2}) = C + t a n^{- 1} x$
(b) $\frac{y}{1 + x^{2}} = C + t a n^{- 1} x$
(c) $y l o g (1 + x^{2}) = C + t a n^{- 1} x$
(d) $y (1 + x^{2}) = C + s i n^{- 1} x$