logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

If $ x=\tan\bigg(\large\frac{1}{a}$$\ log \: y \bigg) $ show that $ (1+x^2)\large\frac{d^2y}{dx^2}$$+(2x-a)\large\frac{dy}{dx}=0 $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\large\frac{d}{dx}$$(\tan^{-1}x)=\large\frac{1}{1+x^2}$
Step 1:
$x=\tan \big(\large\frac{1}{a}$$\log y\big)$
$\tan^{-1}(x)=\large\frac{1}{a}$$\log y$
$a\tan^{-1}(x)=\log y$
Diff w.r.t to $x$ on both sides we get,
$\large\frac{a}{1+x^2}=\large\frac{1}{y}\frac{dy}{dx}$
$\large\frac{ay}{1+x^2}=\frac{dy}{dx}$
$y=\large\frac{dy}{dx}(\frac{1+x^2}{a})$
Step 2:
Diff again w.r.t $x$ we get,
$\large\frac{d^2y}{dx^2}=\frac{(1+x^2).a\large\frac{dy}{dx}-ay(2x)}{(1+x^2)^2}$
$(1+x^2)^2\large\frac{d^2y}{dx^2}$$=a(1+x^2)\large\frac{dy}{dx}$$+2axy$
$(1+x^2)^2\large\frac{d^2y}{dx^2}$$=a(1+x^2)\large\frac{dy}{dx}$$+2ax\large\frac{dy}{dx}(\frac{1+x^2}{a})$
$\large\frac{d^2y}{dx^2}$$(1+x^2)=\large\frac{dy}{dx}$$(a-2x)$
$\large\frac{d^2y}{dx^2}$$(1+x^2)+(2x-a)\large\frac{dy}{dx}$$=0$
Hence proved.
answered Nov 12, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...