The general solution of the nonhomogeneous equation y′′ + 2y′ = 2teᵗ is given by y(t) = c₁e⁻²ᵗ + c₂ + (1/4)t²eᵗ, where c₁ and c₂ are arbitrary constants.
To solve the nonhomogeneous differential equation, we first find the homogeneous solution by setting the right-hand side to zero.
The characteristic equation r² + 2r = 0 has roots r = −2, yielding yₕ(t) = c₁e⁻²ᵗ + c₂. For the particular solution, we use the method of undetermined coefficients.
Assuming yₚ(t) = At²eᵗ + c₂, we find A = 1/4. Therefore, the complete solution is y(t) =yₕ(t) + yₚ(t) = c₁e⁻²ᵗ + c₂ + (1/4)t²eᵗ.
In conclusion, the general solution of the nonhomogeneous equation y′′ + 2y′ = 2teᵗ comprises the homogeneous solution c₁e⁻²ᵗ + c₂ and the particular solution (1/4)t²eᵗ, where c₁ and c₂ are arbitrary constants.
Complete question:
Find the general solution of the nonhomogeneous equation y′′ + 2y′ = 2teᵗ?