Differential Equations
Variable Separable / Partial Fractions
nta_pyq_2024_apr
Grade 12
Question:
If the solution $y=y(x)$ of the differential equation $(x^4+2x^3+3x^2+2x+2)\,dy-(2x^2+2x+3)\,dx=0$ satisfies $y(-1)=-\dfrac{\pi}{4}$, then $y(0)$ is equal to:
$\dfrac{\pi}{2}$
$-\dfrac{\pi}{2}$
$0$
$\dfrac{\pi}{4}$
Step-by-Step Solution
Key Concept: Factor denominator: $x^4+2x^3+3x^2+2x+2=(x^2+1)(x^2+2x+2)$. Partial fractions: $\frac{2x^2+2x+3}{(x^2+1)(x^2+2x+2)}=\frac{1}{x^2+2x+2}+\frac{1}{x^2+1}$.
$y=\tan^{-1}(x+1)+\tan^{-1}x$. $y(0)=\pi/4$.
Correct Answer: 4