Differential Equations
Bernoulli DE — Substitution
nta_pyq_2024_jan
Grade 12
Question:
Let $y=y(x)$ be the solution of the differential equation $\sec^2x\,dx+(e^{2y}\tan^2x+\tan x)\,dy=0$, $0<x<\dfrac{\pi}{2}$, $y\left(\dfrac{\pi}{4}\right)=0$. If $y\left(\dfrac{\pi}{6}\right)=\alpha$, then $e^{8\alpha}$ is equal to
Step-by-Step Solution
Key Concept: Let $t=\tan x$: $\sec^2x\frac{dx}{dy}+e^{2y}t^2+t=0\Rightarrow\frac{dt}{dy}+t=-e^{2y}t^2$ (Bernoulli in $t$). Let $u=1/t$: $\frac{du}{dy}-u=e^{2y}$. Solve linear DE, apply IC.
Let $u=1/\tan x$: $\frac{du}{dy}-u=e^{2y}$. I.F.$=e^{-y}$: $ue^{-y}=e^y+c$. IC $c=0$: $e^{2y}=\cot x$. At $x=\pi/6$: $e^{2\alpha}=\sqrt{3}$. $e^{8\alpha}=3^2=9$.
Correct Answer: 9