Differential Equations
Homogeneous DE — Logarithmic Substitution
nta_pyq_2024_jan
Grade 12
Question:
The solution curve of the differential equation $y\,\dfrac{dx}{dy}=x\left(\log_e x-\log_e y+1\right)$, $x>0$, $y>0$ passing through the point $(e,1)$ is
$\left|\log_e\dfrac{y}{x}\right|=x$
$\left|\log_e\dfrac{y}{x}\right|=y^2$
$\left|\log_e\dfrac{x}{y}\right|=y$
$2\left|\log_e\dfrac{x}{y}\right|=y+1$
Step-by-Step Solution
Key Concept: Rewrite as $\frac{dx}{dy}=\frac{x}{y}\left(\ln\frac{x}{y}+1\right)$. Let $t=\frac{x}{y}$: $x=ty$, $\frac{dx}{dy}=t+y\frac{dt}{dy}$. Separate to get $\frac{dt}{t\ln t}=\frac{dy}{y}$. Integrate using substitution $p=\ln t$.
Let $t=x/y$: $y\frac{dt}{dy}=t\ln t$. $\frac{dt}{t\ln t}=\frac{dy}{y}\Rightarrow\ln(\ln t)=\ln y+c$. IC $(e,1)$: $c=0$. $\ln(x/y)=y\Rightarrow\left|\log_e\frac{x}{y}\right|=y$.
Correct Answer: 3