Answer:
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Explanation:
To decipher to the likely second-order undeviating alike characteristic equating accompanying beginning environments, we can use the form of unproven coefficients.The characteristic equating for the similar constituent the characteristic equating is:r^2 - 4r + 3 = 0Factoring the equating, we have:(r - 1)(r - 3) = 0So the ancestries of the characteristic equating are r1 = 1 and r2 = 3.The inexact resolution for the comparable part is before:y_h(x) = c1 * e^(x) + c2 * e^(3x)To find the indicated resolution, we adopt it has the form:y_p(x) = Ax^2 + Bx + CTaking the descendants of y_p(x), we have:y_p'(x) = 2Ax + By_p''(x) = 2ASubstituting these products into the original characteristic equating, we receive:2A - 4(2Ax + B) + 3(Ax^2 + Bx + C) = 9x^2 - 12Simplifying the equating, we have:(3A)x^2 + (-8A + 3B)x + (2A - 4B + 3C) = 9x^2 - 12Equating the coefficients of like capacities of x, we have the following structure of equatings:3A = 9-8A + 3B = 02A - 4B + 3C = -12Solving this order of equatings, we find A = 3, B = 8, and C = -14.Therefore, the indicated resolution is:y_p(x) = 3x^2 + 8x - 14The common resolution to the original characteristic equating is the total of the alike and particular resolutions:y(x) = y_h(x) + y_p(x) = c1 * e^x + c2 * e^(3x) + 3x^2 + 8x - 14To find the particular resolution accompanying the likely beginning environments, we substitute the principles into the common answer:y(0) = c1 * e^0 + c2 * e^(3*0) + 3(0)^2 + 8(0) - 14 = 6c1 + c2 - 14 = 6y'(x) = c1 * e^x + 3c2 * e^(3x) + 6x + 8y'(0) = c1 * e^0 + 3c2 * e^(3*0) + 6(0) + 8 = 8c1 + 3c2 + 8 = 8Solving this plan of equatings, we find c1 = 10 and c2 = 4.Therefore, the particular resolution to the likely characteristic equating accompanying the primary environments is:y(x) = 10 * e^x + 4 * e^(3x) + 3x^2 + 8x - 14