Differential Equations Question 334
Question: The solution of $ \frac{dy}{dx}+2y\tan x=\sin x $ , is
[DCE 1999]
Options:
A) $ y{{\sec }^{3}}x={{\sec }^{2}}x+c $
B) $ y{{\sec }^{2}}x=\sec x+c $
C) $ y\sin x=\tan x+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{dy}{dx}+2y\tan x=\sin x $ is a linear differential equation of the form $ \frac{dy}{dx}+yf(x)=g(x) $ \ I.F. $ ={e^{\int{f(x)dx}}}={e^{\int{2\tan xdx}}}={e^{2\log (\sec x)}}={e^{\log {{\sec }^{2}}x}}={{\sec }^{2}}x $
Hence, the solution is $ y(I\text{.F}\text{.)}=\int{g(x)I\text{.F}\text{.}dx+c} $
$ y({{\sec }^{2}}x)=\int{\sin x{{\sec }^{2}}xdx+c} $
Therefore $ y{{\sec }^{2}}x=\int{\sec x\tan xdx+c} $
Therefore $ y{{\sec }^{2}}x=\sec x+c $ .
BETA
