Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If y=esin⁡−1x+e−cos⁡−1xy = e^{\sin^{-1}x}+e^{-\cos^{-1}x}, then the value of ∣y′′(x2−1)+yxy′∣\left|\dfrac{y''(x^2-1)+y}{xy'}\right|.
Solution
Answer: 1
Step 1: Simplify yy Using −cos⁡−1x=sin⁡−1x−π2-\cos^{-1}x = \sin^{-1}x - \dfrac{\pi}{2}:
y=esin⁡−1x+esin⁡−1x−π/2=esin⁡−1x ⁣(1+e−π/2)y = e^{\sin^{-1}x}+e^{\sin^{-1}x-\pi/2} = e^{\sin^{-1}x}\!\left(1+e^{-\pi/2}\right)
 Step 2: Derive the differential equation satisfied by yy
y′=y1−x2  ⟹  (1−x2)(y′)2=y2y' = \frac{y}{\sqrt{1-x^2}} \implies (1-x^2)(y')^2 = y^2
 Differentiating and dividing by 2y′2y':
(1−x2)y′′−xy′=y  ⟹  (x2−1)y′′+xy′+y=0(1-x^2)y'' - xy' = y \implies (x^2-1)y'' + xy' + y = 0
 Step 3: Evaluate the expression
(x2−1)y′′+y=−xy′  ⟹  (x2−1)y′′+yxy′=−1(x^2-1)y''+y = -xy' \implies \frac{(x^2-1)y''+y}{xy'} = -1
∣y′′(x2−1)+yxy′∣=1\left|\frac{y''(x^2-1)+y}{xy'}\right| = 1
 Answer: 1
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