Differential Equations
Linear ODE — First Order
nta_pyq_2024_apr
Grade 12
Question:
Let $y=y(x)$ be the solution of the differential equation $(x^2+4)^2\,dy+(2x^3y+8xy-2)\,dx=0$. If $y(0)=0$, then $y(2)$ is equal to:
$\dfrac{\pi}{32}$
$2\pi$
$\dfrac{\pi}{8}$
$\dfrac{\pi}{16}$
Step-by-Step Solution
Key Concept: Rewrite as $\frac{dy}{dx}+y\cdot\frac{2x}{x^2+4}=\frac{2}{(x^2+4)^2}$. IF $=x^2+4$. Solution: $y(x^2+4)=\tan^{-1}(x/2)+C$.
$y(x^2+4)=\tan^{-1}(x/2)$. $y(2)=\pi/(4\cdot8)=\pi/32$.
Correct Answer: 1