The solution of y" = 8(t – 1) – 6(t – 3), y(0) = y'(0) = 0 is y(t) = %3D (A) (t- 3)[u(t - 1) - u(t- 3)] (B) tu(t- 1) - u(t- 3)] (C) (t- 1)[u(t-1) – u(t – 3)] | (D) (t – 1)u(t – 1) – (t – 3)u(t - 3) | (E) (t – 1)²u(t– 1) – (t – 3)²u(t - 3)

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Q: Can iv be explained further.

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Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.