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Q.
Solution of differential equation $\frac{d t}{d x}=\frac{t\left(\frac{d}{d x}(g(x))\right)-t^{2}}{g(x)}$ is
Differential Equations
Solution:
Rearranging the terms of equation, we get
$\frac{d t}{d x}-t \frac{g'(x)}{g(x)}=-\frac{t^{2}}{g(x)}$
$\Rightarrow -\frac{1}{t^{2}} \frac{d t}{d x}+\frac{1}{t} \frac{g^{\prime}(x)}{g(x)}=\frac{1}{g(x)} \dots$(1)
Let $z=\frac{1}{t} \Rightarrow -\frac{1}{t^{2}} \frac{d t}{d x}=\frac{d z}{d x}$
Thus, from (i) we obtain $\frac{d z}{d x}+\frac{g^{\prime}(x)}{g(x)} z=\frac{1}{g(x)}$
which is clearly linear in $z$ and $\frac{d z}{d x}$ with
I.F. $=e^{\int \frac{g'(x)}{g(x)} d x}=e^{\log [g(x)]}=g(x)$
$\Rightarrow $ Thus, complete solution is
$z \cdot g(x)=\int g(x) \cdot \frac{1}{g(x)} d x+C$
$\Rightarrow \frac{1}{t} g(x)=x+c $
$\Rightarrow \frac{g(x)}{x+c}=t$