Toronto Math Forum

y'' -6y' + 25y = 16e^3x + 102sinx
y(0) = 0, y'(0) = 0

(a) The characteristic equation is r² - 6r +25 = 0,
then (r-3)² = -16, so r = 3 + 4i or 3- 4i.

so y_c(x) = c₁e^3xcos4x + c₂e^3xsin4x

Think about y'' -6y' + 25y = 16e^3x.
Let y₁ = Ae^3x,
so y₁' = 3Ae^3x, y₁'' = 9Ae^3x.
Substitute them in the equation,
we get 9Ae^3x - 18Ae^3x +25Ae^3x = 16e^3x.
So 9A-18A+25A = 16, A = 1.
We can get: y₁ = e^3x.

Think about y'' -6y' + 25y = 102sinx.
Let y₂ = Bcosx + Csinx,
so y₂' = -Bsinx +Ccosx,
 y₂'' = -Bcosx - Csinx.
Substitute them in the equation,
we get -Bcosx - Csinx +6Bsinx - 6Ccosx + 25Bcosx + 25Csinx = 102sinx.
So -B - 6C + 25B =0 and -C + 6B + 25C = 102. Then we know B = 1 and C =4.
We can get: y₂ = cosx + 4sinx.

Therefore the general solution is:
y(x) = c₁e^3xcos4x + c₂e^3xsin4x + e^3x + cosx + 4sinx

(b)
y' = c₁(-4e^3xsin4x + 3e^3xcos4x)
+ c₂(4e^3xcos4x + 3e^3xsin4x)
+ 3e^3x - sinx + 4cosx
Substitute  y(0) = 0, y'(0) = 0,
We get 0 = c₁ + 0 + 1 + 1 + 0, so c₁ = -2.
And   0 = c₁(-0 + 3) + c₂(4 + 0) + 3 - 0 + 4
so 0 = -6 + 4c₂ + 7，c₂ = -1/4.

Therefore the solution is:
y(x) = -2e^3xcos4x + (-1/4)e^3xsin4x + e^3x + cosx + 4sinx