Differential Equations
Separable DE — Partial Fractions
nta_pyq_2024_jan
Grade 12
Question:
If $y=y(x)$ is the solution curve of the differential equation $(x^2-4)\,dy-(y^2-3y)\,dx=0$, $x>2$, $y(4)=\dfrac{3}{2}$ and the slope of the curve is never zero, then the value of $y(10)$ equals:
$\dfrac{3}{1+(8)^{1/4}}$
$\dfrac{3}{1+2\sqrt{7}}$
$\dfrac{3}{1-2\sqrt{2}}$
$\dfrac{3}{1-(8)^{1/4}}$
Step-by-Step Solution
Key Concept: Separate: $\frac{dy}{y^2-3y}=\frac{dx}{x^2-4}$. Use partial fractions on both sides. Integrate and apply IC $y(4)=3/2$. Evaluate at $x=10$.
$\frac{1}{3}\ln\left|\frac{y-3}{y}\right|=\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|+C$. IC gives $C=\frac{1}{4}\ln3$. At $x=10$: $\frac{1}{3}\ln\left|\frac{y-3}{y}\right|=\frac{1}{4}\ln\frac{2}{3}+\frac{1}{4}\ln3=\frac{1}{4}\ln2=\ln2^{1/4}$. Since $y\in(0,3)$: $-y+3=8^{1/4}\cdot y\Rightarrow y=\frac{3}{1+8^{1/4}}$.
Correct Answer: 1