Final Answer:
The particular solution to the given differential equation [ y"+6 y'+8 y=-6t e⁵ᵗ ] is yₚ(t) = (1/10)t e⁵ᵗ.
Step-by-step explanation:
To find the particular solution, we assume a particular form for the solution that matches the non-homogeneous term. In this case, we assume a particular solution of the form yₚ(t) = A t e⁵ᵗ, where A is a constant to be determined.
Firstly, calculate the first and second derivatives of yₚ with respect to t:
yₚ(t) = A t e⁵ᵗ
y'ₚ(t) = A e⁵ᵗ + 5A t e⁵ᵗ
y"ₚ(t) = 10A e⁵ᵗ + 25A t e⁵ᵗ
Now, substitute these derivatives into the original differential equation and simplify:
10A e⁵ᵗ + 25A t e⁵ᵗ + 6(A e⁵ᵗ + 5A t e⁵ᵗ) + 8(A t e⁵ᵗ) = -6t e⁵ᵗ
Combine like terms:
10A e⁵ᵗ + 25A t e⁵ᵗ + 6A e⁵ᵗ + 30A t e⁵ᵗ + 8A t e⁵ᵗ = -6t e⁵ᵗ
Combine coefficients of like terms:
16A e⁵ᵗ + 63A t e⁵ᵗ = -6t e⁵ᵗ
Equate coefficients to find A:
16A = -6 \implies A = -3/8
Finally, substitute A back into the assumed particular solution to get the final particular solution:
yₚ(t) = (-3/8) t e⁵ᵗ = (1/10) t e⁵ᵗ