Differential Equations Question 14
Question 14 - 2024 (31 Jan Shift 1)
The solution curve of the differential equation
$y \frac{d x}{d y}=x\left(\log _{e} x-\log _{e} y+1\right), x>0, y>0$ passing
through the point $(e, 1)$ is
(1) $\left|\log _{e} \frac{y}{x}\right|=x$
(2) $\left|\log _{e} \frac{y}{x}\right|=y^{2}$
(3) $\left|\log _{e} \frac{x}{y}\right|=y$
(4) $2\left|\log _{e} \frac{x}{y}\right|=y+1$
Show Answer
Answer (3)
Solution
$\frac{d x}{d y}=\frac{x}{y}\left(\ln \left(\frac{x}{y}\right)+1\right)$
Lat $\frac{x}{y}-t \Rightarrow x-t y$
$\frac{d x}{d y}-t+y \frac{d t}{d y}$
$\mathrm{t}+\mathrm{y} \frac{\mathrm{dt}}{\mathrm{dy}}-\mathrm{t}(\mathrm{l}(\mathrm{t})+1)$
$y \frac{d t}{d y}-t \ln (t) \Rightarrow \frac{d t}{t \ln (t)}=\frac{d y}{y}$
$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t} \cdot \ln (\mathrm{t})}-\int \frac{\mathrm{dy}}{\mathrm{y}}$
$\Rightarrow \int \frac{\mathrm{d}}{\mathrm{p}}-\int \frac{\mathrm{dy}}{\mathrm{y}} \quad \operatorname{let} \ln \mathrm{t}=\mathrm{p}$
$\frac{1}{\mathrm{t}} \mathrm{dt}=\mathrm{dp}$
$\Rightarrow \ln p=\ln y+c$
$\ln (\ln t)=\ln y+c$
$\ln \left(\ln \left(\frac{x}{y}\right)\right)-\ln y+c$
at $\mathbf{x}=\mathbf{e}, y=1$
$\ln \left(\ln \left(\frac{e}{1}\right)\right)-\ln (1)+c \Rightarrow c=0$
$\ln \left|\ln \left(\frac{x}{y}\right)\right|-\ln y$
$\left.\ln \left(\frac{x}{y}\right) \right\rvert,-e^{k y}$
$\left|\ln \left(\frac{x}{y}\right)\right|-y$
