If $x=\tan\!\left(\dfrac{1}{a}\log y\right)$, show that $(1+x^{2})\dfrac{d^{2}y}{dx^{2}}+(2x-a)\dfrac{dy}{dx}=0$.

CBSE Class 12 · 2011 · DifferentiationLong Answer

x = tan (\frac{1}{a} lo g y)

, show that

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

Answer: Proved

From the relation,

\frac{1}{a} lo g y = tan^{- 1} x

, i.e.

lo g y = a tan^{- 1} x

.
Differentiate:

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

Differentiate again (product rule on the left):

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

Transpose:

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

CBSE Class 12 Board · 2011 · Subjective