The general solution of (D²+2D+1)(D²+4)y=0 is A y=(C,+C,x)e+C,cos2x + C ,sin2r B) none of the given choices © y= (C,+C,x)e* +C,cos2x + C ,sin2x -2x +Cze -X E) y=(C,+Cx)e¯+C
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- The solution of some second order linear DEQ is exp(-7x)[5sin(1x)+ 7cos(1x)] which can also be expressed as A*exp(-D*x)*sin(F*x+G). Determine A,D,F, and G (degrees).Obtain the particular solution of (3siny-5x)dx+2x^2cotydy=0 using Bernoulli's Equation. (Complete and Step by step solution please)Suppose that the position of one particle at time t is given by x1 = 3 sin(t), y1 = 2 cos(t), 0 ≤ t ≤ 2? and the position of a second particle is given by x2 = −3 + cos(t), y2 = 1 + sin(t), 0 ≤ t ≤ 2?. (a) Graph the paths of both particles. How many points of intersection are there? __ points of intersection (b) Are any of these points of intersection collision points? That is, are the particles ever at the same place at the same time? If so, find the collision points. (Enter your answers as a comma-separated list of ordered pairs of the form (x, y). If there are no collision points, enter DNE.) (x, y) =__ (c) Describe what happens if the path of the second particle is given by x2 = 3 + cos(t), y2 = 1 + sin(t), 0 ≤ t ≤ 2?. The circle is centered at (x, y) = __ There are _ intersection point(s), and there are _ collision point(s).