Verified answer

general solution of Homogeneous Equation,

propose y = e^(cx)

y” + 10y’ + 25 y = 0

characteristic equation,

m² + 10m + 25 = 0

(m + 5)² = 0

m = -5

homogeneous solution is

y(x) = A e^(-5x) + Bx e^(-5x)

particular solution of Non-Homogeneous Equation,

y = Cx² e^(-5x)

y' = -C e^(-5x)(5x - 2)x

y'' = C e^(-5x) (25x² - 20x + 2)

substitute into the original equation,

y” + 10y’ + 25 y = e^(-5x)

C e^(-5x) (25x² - 20x + 2) - 10 C e^(-5x)(5x - 2)x + 25 Cx² e^(-5x) = e^(-5x)

simplified,

2 C e^(-5x) = e^(-5x)

C = ½

total solution of Non-Homogeneous Equation

y(x) = A e^(-5x) + Bx e^(-5x) + ½ x² e^(-5x)

hence A and B is an arbitrary constant.

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It looks like the above answer has got mixed up, the particular solution means the final solution after the constant has been solved for. It is the particular integral that is one part of the solution which depends on the right hand side on the equation. There is no need to state what is a constant, it is already obvious and goes without saying.

Remember that if the right hand side of the equation appears in the complementary function, then you need to multiply the particular integral by x.

Find the complementary function by using the auxiliary equation:

m² + 10m + 25 = 0

(m + 5)² = 0

m + 5 = 0

m = -5

yᶜ = (Ax + B)℮^(-5x)

yᶜ = (Ax + B) / ℮^(5x)

Find the particular integral by comparing coefficients:

yᵖ = Cx²℮^(-5x)

yᵖ' = 2Cx℮^(-5x) - 5Cx²℮^(-5x)

yᵖ'' = 25Cx²℮^(-5x) - 20Cx℮^(-5x) + 2C℮^(-5x)

yᵖ'' + 10ᵖ' + 25yᵖ = ℮^(-5x)

25Cx²℮^(-5x) - 20Cx℮^(-5x) + 2C℮^(-5x) + 10[2Cx℮^(-5x) - 5Cx²℮^(-5x)] + 25Cx²℮^(-5x) = ℮^(-5x)

25Cx²℮^(-5x) - 20Cx℮^(-5x) + 2C℮^(-5x) + 20Cx℮^(-5x) - 50Cx²℮^(-5x) + 25Cx²℮^(-5x) = ℮^(-5x)

2C℮^(-5x) = ℮^(-5x)

2C = 1

C = ½

yᵖ = x²℮^(-5x) / 2

yᵖ = x² / 2℮^(5x)

Put these together to find the solution to the differential equation:

y = yᶜ + yᵖ

y = (Ax + B) / ℮^(5x) + x² / 2℮^(5x)

y = (x² + Ax + B) / 2℮^(5x)