Differential Equations
Linear ODE — solution involving arcsin
MJAT_TS3_P2
Grade 12
Question:
If $y=f(x)$ is a solution of $\dfrac{d}{dx}\!\left(x^2\dfrac{dy}{dx}\right) + \dfrac{xy}{2} = -\dfrac{1}{x^2\sqrt{1-x^2}}$ and $f(0)=0$, then:
A) $f(1) = \dfrac{\pi}{8}$
B) $f(-1) = -\dfrac{\pi}{8}$
C) $f\!\left(\dfrac{1}{2}\right) = \dfrac{3-2\sqrt{3}}{6}\sin\frac{\pi}{6}$
D) $x^2 f(x) = \dfrac{1}{2}(2x^2-1)\sin^{-1}x + \dfrac{1}{4}x\sqrt{1-x^2} + \dfrac{1}{4}$
Step-by-Step Solution
Key Concept: The ODE (after simplification using IF $= x^2$): $x^2 y = \int x^2 \cdot (\text{RHS})\,dx$. The solution involves $\sin^{-1}x$, giving $x^2 f(x) = \frac{1}{4}(2x^2-1)\sin^{-1}x+\frac{x}{4}\sqrt{1-x^2}-\frac{1}{4}\cdot 0 + C$. With $f(0)=0$: $C=0$.
A ✓, B ✓. Answer: A, B.
Correct Answer: AB