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} = e^2log|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