Differential Equations
Linear DE — Product Rule Recognition
nta_pyq_2024_jan
Grade 12
Question:
Let $y=y(x)$ be the solution of the differential equation $\sec x\,dy+\{2(1-x)\tan x+x(2-x)\}\,dx=0$ such that $y(0)=2$. Then $y(2)$ is equal to:
2
$2\{1-\sin(2)\}$
$2\{\sin(2)+1\}$
1
Step-by-Step Solution
Key Concept: Rewrite as $\frac{dy}{dx}=-\cos x[2(1-x)\tan x+x(2-x)]$. Recognize that $\frac{d}{dx}[(x^2-2x)\sin x]=2(x-1)\sin x+(x^2-2x)\cos x$. The RHS simplifies to $\frac{d}{dx}[(x^2-2x)\sin x]$.
$\frac{dy}{dx}=2(x-1)\sin x-(x^2-2x)\cos x=\frac{d}{dx}[(x^2-2x)\sin x]$... integrating: $y=(x^2-2x)\sin x+c$. $y(0)=2\Rightarrow c=2$. $y(2)=0+2=2$.
Correct Answer: 1