Differential Equations
Exact ODE — Implicit Solution via Separation
nta_pyq_2023_apr
Grade 12
Question:
Let $y=y(x)$ be a solution of $(1-x^2y^2)\,dx=y\,dx+x\,dy$. If $x=1$ gives $y=2$ and $x=2$ gives $y=\alpha$, then a value of $\alpha$ is
$\dfrac{1-3e^2}{2(3e^2+1)}$
$\dfrac{1+3e^2}{2(3e^2-1)}$
$\dfrac{3e^2}{2(3e^2-1)}$
$\dfrac{3e^2}{2(3e^2+1)}$
Step-by-Step Solution
Key Concept: $(1-x^2y^2)dx=d(xy)$. Rearrange: $dx=\frac{d(xy)}{1-(xy)^2}=\frac{1}{2}\!\left(\frac{d(xy)}{1-xy}+\frac{d(xy)}{1+xy}\right)$.
$\left|\frac{xy+1}{xy-1}\right|=3e^{2x-2}$. At $x=2$: $\alpha=\frac{1+3e^2}{2(3e^2-1)}$.
Correct Answer: 2