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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,
yc (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)|
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