Find the general solution of the differential equation given below.

$\frac{dy}{dx} = \frac{1}{x(1 + x^2)}$

$y = \log \left( \frac{cx}{\sqrt{1 + x^2}} \right)$

The correct answer is Option (2) → $y = \log \left( \frac{cx}{\sqrt{1 + x^2}} \right)$ ##

Given differential equation is

$\frac{dy}{dx} = \frac{1}{x(1 + x^2)}$

$\Rightarrow \int dy = \int \frac{dx}{x(1 + x^2)}$

$y = \int \frac{dx}{x(1 + x^2)} \quad \dots(i)$

Now, $\frac{1}{x(1 + x^2)} = \frac{A}{x} + \frac{Bx + C}{1 + x^2} \quad \dots(ii)$

[Using partial fraction]

$1 = A(1 + x^2) + (Bx + C)x$

$\Rightarrow 1 = (A + B)x^2 + Cx + A$

$∴A + B = 0, C = 0 \text{ and } A = 1$

$∴A = 1, B = -1 \text{ and } C = 0$

From Eq. (ii), we get

$\frac{1}{x(1 + x^2)} = \frac{1}{x} - \frac{x}{1 + x^2}$

Now, from Eq. (i), we get

$y = \int \frac{1}{x} dx - \int \frac{x}{1 + x^2} dx$

or, $y = \int \frac{1}{x} dx - \frac{1}{2} \int \frac{2x}{1 + x^2} dx$

or, $y = \log x - \frac{1}{2} \log(1 + x^2) + \log c$

or, $y = \log x - \log \sqrt{1 + x^2} + \log c$

or, $y = \log \left( \frac{cx}{\sqrt{1 + x^2}} \right)$