How did you get “-t-1/2” part? Thankx though!
2 Answers
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General solution= Ke^(-2t)+Ke^t-t-1/2
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Note that the above answer is wrong as there should be two arbitrary constants.
Find the complementary function by solving the auxiliary equation:
y” – y’ – 2 = 0
m² + m – 2 = 0
(m + 2)(m – 1) = 0
m + 2 = 0 OR m – 1 = 0
m = -2 OR m = 1
yᶜ = C₁℮ᵗ + C₂℮^(-2t)
yᶜ = C₁℮ᵗ + C₂ / ℮^(2t)
Find the particular integral by comparing coefficients:
yᵖ = At + B
yᵖ’ = A
yᵖ” = 0
yᵖ” – yᵖ’ – 2yᵖ = 2t
A – 2(At + B) = 2t
A – 2At – 2B = 2t
-2At + (A – 2B) = 2t
-2A = 2
A = -1
A – 2B = 0
2B = A
B = A / 2
B = -½
yᵖ = -t – ½
Find the general solution by combining these two parts:
y = yᶜ + yᵖ
y = C₁℮ᵗ + C₂ / ℮^(2t) – t – ½
