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Find a particular solution to [ y"+6 y'+8 y=-6t e⁵ᵗ ] [ yₚ= ?]

User Dinushan
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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

User Eld
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