Explanation

(c)
$\frac{dt}{dx} - t\frac{g^{'}(x)}{g(x)} = - \frac{t^{2}}{g(x)}$
$\Rightarrow \ - \frac{1}{t^{2}}\frac{dt}{dx} + \frac{1}{t}\ \frac{g^{'}(x)}{g(x)} = \frac{1}{g(x)}$ ….(i)
Let $\mathcal{z =}\frac{1}{t}\ \Rightarrow \ - \frac{1}{t^{2}}\frac{dt}{dx} = \frac{d\mathcal{z}}{dx}$
$\therefore$ From Eq. (i)
$\frac{d\mathcal{z}}{dx} + \frac{g^{'}(x)}{g(x)}\mathcal{z =}\frac{1}{g(x)}$
On comparing with $\frac{d\mathcal{z}}{dx} + P\mathcal{z =}Q$, we get
$P = \frac{g^{'}(x)}{g(x)},\ Q = \frac{1}{g(x)}$
$\therefore IF = e^{\int_{}^{}{\frac{g^{'}(x)}{g(x)}dx}}$
$= e^{\log{\lbrack g(x)\rbrack}} = g(x)$
Thus, complete solution is
$\mathcal{z \bullet}g(x) = \int_{}^{}{g(x) \bullet \frac{1}{g(x)}dx + c}$
$\Rightarrow \frac{1}{t}\ g(x) = x + c\ \Rightarrow \ \frac{g(x)}{x + c} = t$