x log x $\frac{dy}{dx}$ + y = $\frac{2}{x}$ log x Dividing all the terms of the equation by xlogx, we get ⇒ $\frac{dy}{dx} + \frac{y}{x \log x}$ = $\frac{2}{x^2}$ This equation is in the form of a linear differential equation $\frac{dy}{dx}$ + Py = Q , where { = $\frac{1}{x \log x}$ and Q = $\frac{2}{x^2}$ Now, I.F. = $e^{\int P \, dx}$ = $e^{\int \frac{1}{x \log x} \, dx}$ = $e^{\log(\log x)}$ = log x The general solution of the given differential equation is given by y × I.F. = ∫ (Q × I.F.) dx + C ⇒ y log x = ∫ $\left( \frac{2}{x^2} \log x \right)$ dx ⇒ y log x = 2 ∫ $\left( \log x \times \frac{1}{x^2} \right)$ dx = 2 [log x × ∫ $\frac{1}{x^2}$ dx - ∫ $\left\{ \frac{d}{dx} (\log x) \times \int \frac{1}{x^2} \, dx \right\}$ dx] = 2 = 2 $\left[ -\frac{\log x}{x} + \int \frac{1}{x^2} \, dx \right]$ = 2 $\left[ -\frac{\log x}{x} - \frac{1}{x} \right]$ + C So the required general solution is y log x = $-\frac{2}{x}$ (1 + log x) + C OR $\frac{dy}{dx}$ = y tan x ⇒ $\frac{dy}{y}$ = tan x dx On integration, we get ∫ $\frac{dy}{y}$ = ∫ tan x dx ⇒ log y = log (sec x) + log C ... (1) ⇒ log y = log (C sec x) ⇒ y = C sec x Now,it is given that y = 1 when x = 0 ⇒ 1 = C × sec 0 ⇒ 1 = C × 1 ∴ C = 1 Substituting C = 1 in equation (1), we get y = sec x as the required particular solution.