Example: Find a general solution to the subsequent differential equation.

2 y′′ + 18 y + 6 tan (3t)

Solution

First, as the formula for variation of parameters needs coefficients of a one in front of the second derivative let's support that before we not remember. The differential equation which we'll in fact be solving is

y′′ + 9 y + 3 tan (3t)

We'll leave this to you to verify as the complementary solution for that differential equation is,

y_c (t ) = c₁ cos (3t)+ c₂ sin (3t)

 Therefore, we have

 y₁ (t) = cos (3t)

 y₂ (t) = sin (3t)

 The Wronskian of these two functions is

= 3cos²(3t) + 3sin²(3t) = 3

The particular solution is after that,

The general solution is,

y (t ) = c1 cos (3t ) + c2 sin(3t) In |sec(3t) + tan (3t)|