Find y as a function of t if 100y''-100y'+29y=0 y(0)=4, y'(0)=8

100y''-100y'+29y=0
100r^2 - 100r + 29 = 0
r^2 - r + 29/100 = 0
r^2 - r = - 29/100
r^2 - r + 1/4 = - 29/ 100 + 1/4
(r - 1/2)^2 = - 1/25
r - 1/2 = ± √(-1/25)
r = 1/2 ± i/5

general solution

y = e^(1/2t)[c₁sin(t/5) + c₂cos(t/5)]
y = c₁e^(t/2)sin(t/5) + c₂e^(t/2)cos(t/5)

when y(0) = 4

4 = c₁e^[1/2(0)]sin(0/5) + c₂e^[1/2(0)]cos(0/5)
4 = c₂

when y'(0) = 8

y = c₁e^(1/2t)sin(t/5) + c₂e^(1/2t)cos(t/5)
y' = c₁[sin(t/5) d/dt 1/2e^(1/2t) + e^(1/2t) d/dt 1/5cos(t/5)]
+ c₂[cos(t/5) d/dt 1/2e^(1/2t) + e^(1/2t) d/dt - 1/5sin(t/5)

y' = c₁[1/2e^(t/2)sin(t/5) + 1/5e^(t/2)cos(t/5) + c₂ 1/2e^(t/2)cos(t/5) - 1/5e^(t/2)sin(t/5)
y' = 1/2c₁e^(t/2)sin(t/5) + 1/5c₁e^(t/2)cos(t/5) + 1/2c₂e^(t/2)cos(t/5) - 1/5c₂e^(t/2)sin(t/5)
8 = 1/2c₁e^[1/2(0)]sin(0/5) + 1/5c₁e^[1/2(0)]cos(0/5) + 1/2c₂e^[1/2(0)]cos(0/5) - 1/5c₂e^[1/2(0)]sin(0/5)
8 = 1/5c₁ + 1/2c₂ eq2

since c₂ = 4

8 = 1/5c₁ + 1/2(4)
8 = 1/5c₁ + 2
8 - 2 = 1/5c₁
6 = 1/5c₁
c₁ = 30

therefore, the general solution is

y = 30e^(t/2)sin(t/5) + 4e^(t/2)cos(t/5)
y = 2e^(t/2)[15sin(t/5) + 2cos(t/5)] answer//