If $y=\log\!\left[x+\sqrt{x^{2}+a^{2}}\right]$, show that $(x^{2}+a^{2})\dfrac{d^{2}y}{dx^{2}}+x\dfrac{dy}{dx}=0$.

CBSE Class 12 · 2013 · DifferentiationLong Answer

y = lo g [x + x^{2} + a^{2}]

, show that

(x^{2} + a^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x} = 0

Answer: Proved

Differentiate once. Since

\frac{d}{d x} (x + x^{2} + a^{2}) = 1 + \frac{x}{x ^{2} + a ^{2}} = \frac{x + x ^{2} + a ^{2}}{x ^{2} + a ^{2}}

\frac{d y}{d x} = \frac{1}{x ^{2} + a ^{2}} \Rightarrow x^{2} + a^{2} \frac{d y}{d x} = 1

Square:

(x^{2} + a^{2}) (\frac{d y}{d x})^{2} = 1

. Differentiate again:

2 x (\frac{d y}{d x})^{2} + (x^{2} + a^{2}) \cdot 2 \frac{d y}{d x} \frac{d ^{2} y}{d x ^{2}} = 0

Divide by

2 \frac{d y}{d x}

(x^{2} + a^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x} = 0

CBSE Class 12 Board · 2013 · Subjective