Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Solve : $y' +\tan x.y =\tan x \quad y(\large\frac{\pi}{4})$$=1$

$(a)\;y= e^{\sec ^2 x}\frac{\tan ^2x}{2}+c \\ (b)\;y= e^{\tan ^2 x/2}+c \\ (c)\;y= e^{\sec ^2 x}\frac{\tan ^2x}{4}+c \\ (d)\;y= e^{-\sec ^2 x}\frac{\tan ^2x}{2}+c $

Can you answer this question?

1 Answer

0 votes
It is linear equation so
$I_0 f=e^{\int \tan xdx}$
$\qquad= e^{\sec^2 x dx}$
So, $y. e^{\sec^2 x}=\int \tan x . \sec^2 x dx$
$\tan x =t$
$\sec^2 x dx =dt$
$y. e^{\sec^2 x}=\int t dt$
$y. e^{\sec^2 x} =\large\frac{\tan ^2 x}{2}$$+c$
Hence d is the correct answer.
answered Feb 6, 2014 by meena.p

Related questions

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