According to the method of undetermined coefficients, the differential equation y" — y' — 6 y = 2 t² + 3 te³t - e2t will have a particular solution in the form - Ур C1 C2t+C3t² + C4 e³t + С5 e 2t = Y₁ = C₁+ C₂t + C3t² + (C4 + C5 t) e³t + (C 6 + C7 t) e²t Оур 2t Ур C₁ t² + C₂ te³ + C³ e²± Ур о ур == = = C₁ + C₂ t + C3 t² + (C₁t + C5 ±²) e³t + C6 e² 2t : C₁ t² + C₂t + C3 + (C₁ t + С5) e³t + C6 e²± Which of the following is the solution of the initial value problem О О ft²y" +4ty' + 2y = 0 |\y(1) = 2, y′(1) = −1 y = 3 t¹ — t−2 - О О У y=t+t-2 y = 3 t² — t−1 = 5 3 ť -1 - + 1 3 t2 y = 5t-3t²