Question

If log y = tan^–1 X, show that : (1+x²)y₂+(2x–1) y₁=0.

Answer 1

Note: y₂ represents second order derivative i.e.\(\frac{d^2y}{dx^2}\) and y₁ = dy/dx

Given,

log y = tan^–1 X

y = \(e^{tan^{-1}x}\)...............equation 1

to prove : (1 - x²)y₂ -xy₁ - a²=0

We notice a second–order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find \(\frac{d^2y}{dx^2}\)

As, \(\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})\)

So, lets first find dy/dx

\(\frac{dy}{dx}=\frac{d}{dx}e^{tan^{-1}x}\)

Using chain rule, we will differentiate the above expression

Let t = tan –1 x => \(\frac{dt}{dx}\)=\(\frac{1}{1+x^2}\)[ \(\frac{d}{dx}tan^{-1}x=\frac{1}{1+x^2}\) ]

And y = e^t

\(\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}\)

\(\frac{dy}{dx}=e^t\frac{1}{1+x^2}=\frac{e^{tan^{-1}x}}{1+x^2}\)...........equation 2

Again differentiating with respect to x applying product rule:

Using chain rule we will differentiate the above expression-

Using equation 2 :

∴ (1+x²)y₂+(2x–1)y₁=0 ……proved