Solve the IVP y6y8y3sin2t y00 y00 ytSolutionGiven y6y8y3sin2

Solve the IVP

y\'\'+6y\'+8y=-3sin(2t)

y(0)=0

y\'(0)=0

y(t)=?

Solution

Given

y\'\'+6y\'+8y=-3sin(2t)

converting this trigonometric form to exponential form

we know sint = e^(it) - e^(-it) / 2i

sin(2t) = e^(2it) - e^(-2it) / 2i

Y\'\' + 6Y\' + 8Y=0

r^2 + 6r+8 =0

r^2 + 4r +2r +8=0

r(r+4) + 2(r+4) =0

(r+2) (r+4)=0

r= -2 and r=-4

y(t) = C1 e^(-4it) + C2 e^(-8it)

= A e^(-4it) + Bt e^(-8it)

= A e^(-4it) +Be^(-8it) - Bt e^(-8it)

= A e^(-4it) -Be^(-8it) -Be^(-8it) - Bt e^(-8it)

= A e^(-4it) -2 Be^(-8it) - Bt e^(-8it)

=  A e^(-4it) -2 Be^(-8it) + Bt e^(-8it) -A e^(-4it) -Be^(-4it) + Bte^(-8it) -2Ae^(-4it) -2Bte^(-8it)

= -2Ae^(-4it) - 3Be^(-8it)

-2Ae^(-4it) - 3Be^(-8it) = 3(e^(-4it)/2 + e^(-8it)/2)

comparing both sides we get

A = -3/4 and B = -3/6

from here y(t) = -3e^(-4it)/4 -3e^(-8it)/6

hence this is the required solution.