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