Differential Equations
Differential Equations
nta_pyq_2025_apr
Grade 12
Question:
Let $y = f(x)$ be the solution of the differential equation $\dfrac{dy}{dx}+\dfrac{xy}{x^2-1} = \dfrac{x^6+4x}{\sqrt{1-x^2}}$, $-1<x<1$ such that $f(0) = 0$. If $6\displaystyle\int_{-1/2}^{1/2}f(x)\,dx = 2\pi-\alpha$ then $\alpha^2$ is equal to ____.
Step-by-Step Solution
Key Concept: Rewrite $\tfrac{x}{x^2-1} = -\tfrac{x}{1-x^2}$; I.F. $= e^{\frac{1}{2}\ln(1-x^2)} = \sqrt{1-x^2}$; integrate $y\sqrt{1-x^2} = \int(x^6+4x)dx$ and use $f(0)=0$ to find $C=0$. Then compute the integral using symmetry and $x=\sin\theta$.
$\dfrac{dy}{dx}-\dfrac{x}{1-x^2}y = \dfrac{x^6+4x}{\sqrt{1-x^2}}$. I.F. $= \sqrt{1-x^2}$.
$y\sqrt{1-x^2} = \dfrac{x^7}{7}+2x^2+C$. At $x=0$: $C=0$.
$6\int_{-1/2}^{1/2}f(x)dx = 6\int_{-1/2}^{1/2}\dfrac{x^7/7+2x^2}{\sqrt{1-x^2}}dx = 24\int_0^{1/2}\dfrac{x^2}{\sqrt{1-x^2}}dx$ (odd terms vanish).
Let $x=\sin\theta$: $= 24\int_0^{\pi/6}\sin^2\theta\,d\theta = 12\left[\theta-\tfrac{\sin 2\theta}{2}\right]_0^{\pi/6} = 12\left(\tfrac{\pi}{6}-\tfrac{\sqrt{3}}{4}\right) = 2\pi-3\sqrt{3}$.
So $\alpha = 3\sqrt{3}$ and $\alpha^2 = 27$.
Correct Answer: 27