Question

Find a particular solution satisfying the given condition: $\frac {dy}{dx}+2y\ tan\ x=sin\ x;\ y=0\ when \ x=\frac \pi3$

Answer 1

The given differential equation is $\frac {dy}{dx}$+2y tan x = sin x.

This is a linear equation of the form:

$\frac {dy}{dx}$+py = Q (where p=2 tan x and Q=sin x)

Now, I.F. = e∫pdx = e∫2tan xdx = e2log|secx| = $e^{log(sec^2x)}$= sec2x

The general solution of the given differential equation is given by the relation,

y(I.F.) = ∫(Q×I.F.)dx + C

⇒y (sec2x) = ∫(sin x . sec2x)dx + C

⇒ysec2x = ∫(sec x . tan x)dx + C

⇒y sec2x = sec x + C …….....(1)

Now, y=0 at x=$\frac \pi3$

Therefore,

0×sec2$\frac \pi3$ = sec$\frac \pi3$+C

⇒0 = 2+C

⇒C = -2

Substituting C=-2 in equation(1), we get:

y sec2x = sec x - 2

⇒y = cos x - 2cos2x

Hence,the required solution of the given differential equation is y = cos x - 2cos2x