Question

Solution of differential equation $\frac{dt}{dx}=\frac{t\left(\frac{d}{dx}(g(x))\right)-t^2}{g(x)}$ is

• A. $t=\frac{g(x)}{x}+c$
• B. $t=\frac{g(x)}{x^2}+c$
• C. $t=\frac{g(x)}{x+c}$
• D. $t=g(x)+x+c$

Answer 1

Answer: (C)

Rearranging the terms of equation, we get
$\frac{dt}{dx}-t\frac{g^{\prime }(x)}{g(x)}=-\frac{t^2}{g(x)}$
$\Rightarrow -\frac{1}{t^2}\frac{dt}{dx}+\frac{1}{t}\frac{g^{\prime }(x)}{g(x)}=\frac{1}{g(x)}\dots$(1)
Let $z=\frac{1}{t}\Rightarrow -\frac{1}{t^2}\frac{dt}{dx}=\frac{dz}{dx}$
Thus, from (i) we obtain $\frac{dz}{dx}+\frac{g^{\prime }(x)}{g(x)}z=\frac{1}{g(x)}$
which is clearly linear in $z$ and $\frac{dz}{dx}$ with
I.F. $=e^{\int \frac{g^{\prime }(x)}{g(x)}dx}=e^{\log [g(x)]}=g(x)$
$\Rightarrow$ Thus, complete solution is
$z\cdot g(x)=\int g(x)\cdot \frac{1}{g(x)}dx+C$
$\Rightarrow \frac{1}{t}g(x)=x+c$
$\Rightarrow \frac{g(x)}{x+c}=t$