If y = [ x + \sqrt x^2 - 1 ]^15 + [ x - \sqrt x^2 - 1 ]^15, then (x2 - 1) d^2y \over dx^2 + xdy \over dx is equal to :

Differentiation PYQ — JEE Main 2017 · 8 Apr · Shift 1 · Q47 with Solution

View written solutionFree

Correct answer: D

Let $a=x+\sqrt{x^2-1},\qquad b=x-\sqrt{x^2-1}.$ Then $ab=(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})=1.$ Also, $y=a^{15}+b^{15}.$
Use the standard substitution $x=\cosh t \quad (|x|\ge 1),$ so that $\sqrt{x^2-1}=\sinh t.$ Hence $a=\cosh t+\sinh t=e^t,\qquad b=\cosh t-\sinh t=e^{-t}.$ Therefore $y=e^{15t}+e^{-15t}=2\cosh(15t).$ Also, $x=\cosh t.$
Differentiate with respect to $t$ : $\frac{dy}{dt}=30\sinh(15t),$ $\frac{dx}{dt}=\sinh t.$ So $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{30\sinh(15t)}{\sinh t}.$
Now compute the required expression $(x^2-1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}.$ A useful identity is: if $x=\cosh t$ , then $x^2-1=\sinh^2 t,$ and $\frac{d}{dx}=\frac{1}{\sinh t}\frac{d}{dt}.$ Thus $\frac{dy}{dx}=\frac{1}{\sinh t}\frac{dy}{dt}.$ Then

=\sinh^2 t\cdot \frac{d}{dx}\left(\frac{1}{\sinh t}\frac{dy}{dt}\right)+\cosh t\cdot \frac{1}{\sinh t}\frac{dy}{dt}. $$ Since $$\frac{d}{dx}=\frac{1}{\sinh t}\frac{d}{dt},$$ we get $$\sinh^2 t\cdot \frac{1}{\sinh t}\frac{d}{dt}\left(\frac{1}{\sinh t}\frac{dy}{dt}\right)+\cosh t\cdot \frac{1}{\sinh t}\frac{dy}{dt}$$ $$=\sinh t\frac{d}{dt}\left(\frac{1}{\sinh t}\frac{dy}{dt}\right)+\frac{\cosh t}{\sinh t}\frac{dy}{dt}.$$ Expanding the derivative, $$\sinh t\left(-\frac{\cosh t}{\sinh^2 t}\frac{dy}{dt}+\frac{1}{\sinh t}\frac{d^2y}{dt^2}\right)+\frac{\cosh t}{\sinh t}\frac{dy}{dt}$$ $$=-\frac{\cosh t}{\sinh t}\frac{dy}{dt}+\frac{d^2y}{dt^2}+\frac{\cosh t}{\sinh t}\frac{dy}{dt}$$ $$=\frac{d^2y}{dt^2}.$$ So the whole expression simplifies to $$ (x^2-1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=\frac{d^2y}{dt^2}. $$ 5. Now, $$y=2\cosh(15t),$$ so $$\frac{d^2y}{dt^2}=2\cdot 15^2\cosh(15t)=225\cdot 2\cosh(15t)=225y.$$ 6. Therefore, $$ (x^2-1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=225y. $$ 7. Checking options: - A: $125y$ ❌ - B: $124y^2$ ❌ - C: $225y^2$ ❌ - D: $225y$ ✅ Hence the correct answer is **Option D**.

View written solutionFree

Correct answer: D

Let $a=x+\sqrt{x^2-1},\qquad b=x-\sqrt{x^2-1}.$ Then $ab=(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})=1.$ Also, $y=a^{15}+b^{15}.$
Use the standard substitution $x=\cosh t \quad (|x|\ge 1),$ so that $\sqrt{x^2-1}=\sinh t.$ Hence $a=\cosh t+\sinh t=e^t,\qquad b=\cosh t-\sinh t=e^{-t}.$ Therefore $y=e^{15t}+e^{-15t}=2\cosh(15t).$ Also, $x=\cosh t.$
Differentiate with respect to $t$ : $\frac{dy}{dt}=30\sinh(15t),$ $\frac{dx}{dt}=\sinh t.$ So $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{30\sinh(15t)}{\sinh t}.$
Now compute the required expression $(x^2-1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}.$ A useful identity is: if $x=\cosh t$ , then $x^2-1=\sinh^2 t,$ and $\frac{d}{dx}=\frac{1}{\sinh t}\frac{d}{dt}.$ Thus $\frac{dy}{dx}=\frac{1}{\sinh t}\frac{dy}{dt}.$ Then

=\sinh^2 t\cdot \frac{d}{dx}\left(\frac{1}{\sinh t}\frac{dy}{dt}\right)+\cosh t\cdot \frac{1}{\sinh t}\frac{dy}{dt}. $$ Since $$\frac{d}{dx}=\frac{1}{\sinh t}\frac{d}{dt},$$ we get $$\sinh^2 t\cdot \frac{1}{\sinh t}\frac{d}{dt}\left(\frac{1}{\sinh t}\frac{dy}{dt}\right)+\cosh t\cdot \frac{1}{\sinh t}\frac{dy}{dt}$$ $$=\sinh t\frac{d}{dt}\left(\frac{1}{\sinh t}\frac{dy}{dt}\right)+\frac{\cosh t}{\sinh t}\frac{dy}{dt}.$$ Expanding the derivative, $$\sinh t\left(-\frac{\cosh t}{\sinh^2 t}\frac{dy}{dt}+\frac{1}{\sinh t}\frac{d^2y}{dt^2}\right)+\frac{\cosh t}{\sinh t}\frac{dy}{dt}$$ $$=-\frac{\cosh t}{\sinh t}\frac{dy}{dt}+\frac{d^2y}{dt^2}+\frac{\cosh t}{\sinh t}\frac{dy}{dt}$$ $$=\frac{d^2y}{dt^2}.$$ So the whole expression simplifies to $$ (x^2-1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=\frac{d^2y}{dt^2}. $$ 5. Now, $$y=2\cosh(15t),$$ so $$\frac{d^2y}{dt^2}=2\cdot 15^2\cosh(15t)=225\cdot 2\cosh(15t)=225y.$$ 6. Therefore, $$ (x^2-1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=225y. $$ 7. Checking options: - A: $125y$ ❌ - B: $124y^2$ ❌ - C: $225y^2$ ❌ - D: $225y$ ✅ Hence the correct answer is **Option D**.