Question: solve the initial value problem y"+3y'+2.25y=-10e^(-1.5x) if y(0)=1, y'(0)=0.
Auxiliary equation
m^2 + 3m + 2.25 = 0
(m + 1.5)^2 = 0
m = -1.5 (twice - a repeated root)
Complementary solution
y1 = exp(-1.5x)(A + Bx)
Particular solution
Since the rhs of the DE = -10exp(-1.5x) and the complementary solution already has a exp(-1.5x) and x.exp(-1.5x) as part of it, then assume a particular solution of the form y2 = Cx^2.exp(-1.5x)
Then, y2' = 2Cx.exp(-1.5x) - 1.5Cx^2.exp(-1.5x),
and y2'' = 2C.exp(-1.5x) -3.0Cx.exp(-1.5x) - 3.0Cx.exp(-1.5x) + 2.25Cx^2.exp(-1.5x)
i.e. y2'' = 2C.exp(-1.5x) - 6.0Cx.exp(-1.5x) + 2.25Cx^2.exp(-1.5x)
Substituting for y2, y2' and y2'' into the DE,
y2" + 3y2' + 2.25y2 = -10exp(-1.5x)
2C.exp(-1.5x) - 6.0Cx.exp(-1.5x) + 2.25Cx^2.exp(-1.5x) + 3(2Cx.exp(-1.5x) - 1.5Cx^2.exp(-1.5x)) + 2.25(Cx^2.exp(-1.5x)) = -10exp(-1.5x)
2A - 6.0Cx + 2.25Cx^2 + 6Cx - 4.5Cx^2 + 2.25Cx^2 = -10
2C + (-6.0C + 6C)x + (2.25C - 4.5C + 2.25C)x^2 = -10
2C = -10
C = -5
Hence the particular solution is,
y2 = -5x^2.exp(-1.5x)
General Solution
y = y1 + y2
y(x) = exp(-1.5x)(A + Bx) - 5x^2exp(-1.5x)
y(x) = exp(-1.5x)(A + Bx - 5x^2)
Initial conditions
y(0)=1, y'(0)=0.
and y'(x) = -1.5.exp(-1.5x)(A + Bx - 5x^2) + exp(-1.5x)(B - 10x)
y'(x) = exp(-1.5)(-1.5A - 1.5Bx + 7.5x^2 + B - 10x)
y'(x) = exp(-1.5x)(B - 1.5A - (10 + 1.5B)x + 7.5x^2)
At x = 0, y(0) = 1 = exp(0)(A + 0 - 0) = A --> A = 1
At x = 0, y'(0) = 0 = exp(0)(B - 1.5A - 0 + 0) = B - 1.5A = B - 1.5 --> B = 1.5
Hence, A = 1, B = 1.5
Final solution
y(x) = exp(-1.5x)(1 + 1.5x - 5x^2)