Solve each of the following initial value problems:
, y = 2 when
, y = 2 when
Given and
This is a first order linear differential equation of the form
Here, P = –3 cot x and Q = sin 2x
The integrating factor (I.F) of this differential equation is,
We have
[∵ m log a = log am]
∴ I.F = cosec3x [∵ elog x = x]
Hence, the solution of the differential equation is,
Recall
⇒ ycosec3x = 2(–cosec x) + c
⇒ ycosec3x = –2cosec x + c
∴ y = –2sin2x + csin3x
However, when, we have y = 2
⇒ 2 = –2(1)2 + c(1)3
⇒ 2 = –2 + c
∴ c = 4
By substituting the value of c in the equation for y, we get
y = –2sin2x + (4)sin3x
∴ y = –2sin2x + 4sin3x
Thus, the solution of the given initial value problem is y = –2sin2x + 4sin3x