Question

Find a particular solution satisfying the given condition:$\frac {dy}{dx}-3y\ cot\ x=sin2x;\ y=2 \ when\ x=\frac \pi2$

Answer 1

The differential equation is $\frac {dy}{dx}$3y cot x = sin 2x

This is a linear differential equation of the form:

$\frac {dy}{dx}$ + py = Q (where p=-3 cot x and Q=sin 2x)

Now, I.F. = ∫(Q×I.F.)dx + C

⇒y.$\frac {1}{sin^3x}$=∫[sin2x.$\frac {1}{sin^3x}$]dx + C

⇒$\frac {y}{sin ^3x}$ = 2∫(cot x . cosec x)dx + C

⇒$\frac {y}{sin ^3x}$ = -2 cosec x + c

⇒y = -2sin2x + C sin3x ……....(1)

Now, y=2 at x=$\frac \pi2$

Therefore,we get:

2=-2+C

⇒C=4

Substituting C=4 in equation(1),we get:

y = -2sin2x + 4sin3x

⇒y = 4sin3x - 2sin2x

This is the required particular solution of the given differential equation.