CUET UG 2025 — Mathematics Algebra

Given $y = e^{a\cos^{-1}x}$y=eacos−1x

Using the chain rule, and recalling that $\dfrac{d}{dx}(\cos^{-1}x) = \dfrac{-1}{\sqrt{1-x^2}}$dxd​(cos−1x)=1−x2​−1​:

$$\dfrac{dy}{dx} = e^{a\cos^{-1}x} \cdot a \cdot \left(\dfrac{-1}{\sqrt{1-x^2}}\right) = \dfrac{-a \cdot y}{\sqrt{1-x^2}}$$dxdy​=eacos−1x⋅a⋅(1−x2​−1​)=1−x2​−a⋅y​

---

Multiplying both sides by $\sqrt{1-x^2}$1−x2​:

$$\sqrt{1-x^2}\;\dfrac{dy}{dx} = -ay$$1−x2​dxdy​=−ay

Squaring both sides:

$$(1-x^2)\left(\dfrac{dy}{dx}\right)^2 = a^2y^2 \quad \cdots (i)$$(1−x2)(dxdy​)2=a2y2⋯(i)

---

Differentiating equation $(i)$(i) with respect to $x$x:

$$(-2x)\left(\dfrac{dy}{dx}\right)^2 + (1-x^2)\cdot 2\dfrac{dy}{dx}\cdot\dfrac{d^2y}{dx^2} = 2a^2 y\cdot\dfrac{dy}{dx}$$(−2x)(dxdy​)2+(1−x2)⋅2dxdy​⋅dx2d2y​=2a2y⋅dxdy​

---

Since $y = e^{a\cos^{-1}x} > 0$y=eacos−1x>0 always, $\dfrac{dy}{dx} \neq 0$dxdy​=0, so dividing both sides by $2\dfrac{dy}{dx}$2dxdy​:

$$-x\dfrac{dy}{dx} + (1-x^2)\dfrac{d^2y}{dx^2} = a^2 y$$−xdxdy​+(1−x2)dx2d2y​=a2y

$$(1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} = a^2 y$$(1−x2)dx2d2y​−xdxdy​=a2y