Differential Equations
Linear ODE — Finding Value at Specific Point
nta_pyq_2023_apr
Grade 12
Question:
If the solution curve of $(y-2\ln x)\,dx+(x\ln x^2)\,dy=0$, $x>1$ passes through $(e,\frac{4}{3})$ and $(e^4,\alpha)$, then $\alpha$ is equal to _______.
Step-by-Step Solution
Key Concept: Rewrite: $2x\ln x\,\frac{dy}{dx}+y=2\ln x$. Divide by $2x\ln x$: $\frac{dy}{dx}+\frac{y}{2x\ln x}=\frac{1}{x}$. IF $=\sqrt{\ln x}$.
$y\sqrt{\ln x}=\frac{2}{3}(\ln x)^{3/2}+\frac{2}{3}$. At $x=e^4$: $\alpha=3$.
Correct Answer: 3