Differential Equations
Differential Equations
nta_pyq_2025_apr
Grade 12
Question:
Let $y = y(x)$ be the solution of the differential equation $\cos x\,(\log_e(\cos x))^2\,dy + (\sin x - 3y\sin x\log_e(\cos x))\,dx = 0$, $x\in\left(0,\dfrac{\pi}{2}\right)$. If $y\!\left(\dfrac{\pi}{4}\right) = \dfrac{-1}{\log_e 2}$, then $y\!\left(\dfrac{\pi}{6}\right)$ is equal to:
$\dfrac{1}{\log_e(3)-\log_e(4)}$
$\dfrac{-2}{\log_e(3)-\log_e(4)}$
$\dfrac{1}{\log_e(4)-\log_e(3)}$
$\dfrac{-1}{\log_e(4)}$
Step-by-Step Solution
Key Concept: Let $t = \ln(\cos x)$; then $-\tan x\,dx = dt$, and the ODE reduces to $t^2 dy + (-3ty)(-dt) = (-dt)$ ... leading to a linear ODE in $y(t)$ with I.F. $= t^3 = (\ln\cos x)^3$.
Rearranging: $\dfrac{dy}{dx} - \dfrac{3\sin x}{\cos x\ln(\cos x)}y = \dfrac{-\sin x}{\cos x(\ln\cos x)^2}$.
Let $t=\ln(\cos x)$, $dt = -\tan x\,dx$. I.F. $= e^{3\int dt/t} = t^3 = (\ln\cos x)^3$.
$y(\ln\cos x)^3 = \dfrac{(\ln\cos x)^2}{2}+C$.
At $x=\pi/4$: $y(-\ln 2)^3 = \dfrac{(-\ln 2)^2}{2}+C$ and $y(\pi/4) = -1/\ln 2$:
$(-1/\ln 2)(-\ln 2)^3 = \dfrac{(\ln 2)^2}{2}+C \Rightarrow (\ln 2)^2 = \dfrac{(\ln 2)^2}{2}+C \Rightarrow C=0$.
$y = \dfrac{1}{2\ln(\cos x)}$.
$$y\!\left(\frac{\pi}{6}\right) = \frac{1}{2\ln(\sqrt{3}/2)} = \frac{1}{\ln 3-\ln 4}.$$
Correct Answer: 1