Final answer:
To solve the initial-value problem, first find the complementary function and the particular solution of the non-homogeneous equation. Then, combine them to find the solution to the initial-value problem. The solution is y = c₁ * e^(1/2x) + c2 * e⁻²ˣ + Ax₂ + Bx + C.
Step-by-step explanation:
To solve the initial-value problem, we first need to find the complementary function of the given homogeneous equation, which is 2y'' + 3y' - 2y = 0.
The characteristic equation of this homogeneous equation is 2r² + 3r - 2 = 0, where r represents the roots.
Solving this quadratic equation gives us the roots r = 1/2 and r = -2.
Therefore, the complementary function is y_c = c₁ * e^(1/2x) + c₂ * e⁻²ˣ, where c₁ and c₂ are constants.
Next, we find the particular solution of the non-homogeneous equation 2y'' + 3y' - 2y = 18x² - 8x - 15 using the method of undetermined coefficients.
Assuming the particular solution has the form y_p = Ax² + Bx + C, we substitute it into the equation and solve for A, B, and C.
Finally, the solution to the initial-value problem is y = y_c + y_p = c1 * e^(1/2x) + c₂ * e⁻²ˣ + Ax² + Bx + C, where c₁, c₂, A, B, and C are determined by the initial conditions y(0) = 0 and y'(0) = 0.