The general solution of y'' + y = 0 is reached by solving λ2 + 1 = 0, which has roots λ= ±i where i = √-1. Complex Mathematics pioneered by Leonhard Euler uses these roots to give the complementary solution to y'' + y = 0 as yh = C1cos t + C2sin t.
The general solution then consists of all linear combinations of y1 = cos t and y2 = sin t. To solve by Variation Of Parameters, first rewrite 3sin 2t + tcos 2t as 6(sin t)cos t + tcos2t - tsin2t.
Next, calculate The Wronskian of y1 and y2: W(y1,y2) = y1y2' - y1'y2 or cos2t - (-sin2t) or 1. Next write v1 = -∫y2(6(sin t)cos t + tcos2t-tsin2t)dt/W(y1,y2) or -∫(6sin2t(cos t)+tcos2t(sin t) - tsin3t)dt. Using several Trigonometric Identities and Integration By Parts, v1 can be obtained as -2sin3t - tcos t+(2/3)tcos3t+(1/3)sin t+(2/9)sin3t.
Now write v2 =∫y1(6(sin t)cos t + tcos2t - tsin2t)dt/W(y1,y2) or ∫(6(sin t)cos2t + tcos3t - t(sin2t)cos t)dt which will eventually transform by Trigonometric Identities and Integration By Parts to -2cos3t+tsin t-(2/3)tsin3t+(1/3)cos t+(2/9)cos3t.
The final task is to evaluate y1v1 + y2v2:
y1v1 equals -2sin3t(cos t) - tcos2t + (2/3)tcos4t +(1/3)(sin t)cos t +(2/9)sin3t(cos t).
y2v2 equals -2cos3t(sin t) + tsin2t - (2/3)tsin4t +(1/3)(cos t)sin t +(2/9)cos3t(sin t).
A considerable amount of Algebraic and Trigonometric manipulation will reduce the sum of the two "strings" for y1v1 and y2v2 written above to (-1+1/3+1/9)sin 2t + (-1+2/3)tcos 2t equal to -(5/9)sin 2t -(1/3)tcos 2t. This is the Particular Solution, designated by yp.
The general solution sought is then reached by adding yh to yp, which gives C1cos t + C2sin t -(1/3)tcos 2t - (5/9)sin 2t.
