Final answer:
To solve the differential equation y(⁴) - 4y'' = t² + eᵗ, find the complementary solution for the homogeneous part first, and then find a particular solution for the non-homogeneous part. The general solution is the sum of these two solutions.
Step-by-step explanation:
To determine the general solution of the differential equation y(⁴) - 4y'' = t² + eᵗ, we first address the homogeneous part of the equation, which is y(⁴) - 4y'' = 0. This is a linear homogeneous equation with constant coefficients. The corresponding characteristic equation is r⁴ - 4r² = 0, which simplifies to r²(r² - 4) = 0, yielding roots r = 0, 0, 2, -2. Therefore, the complementary solution (or homogeneous solution) is yc = C1 + C2t + C3e2t + C4e-2t.
Next, we need to find a particular solution, yp, for the non-homogeneous part t² + eᵗ. We assume a form for yp based on the terms of the non-homogeneous part and solve for the unknown coefficients.
Finally, the general solution is the sum of the complementary and particular solutions: y = yc + yp.
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