MATH 1B Lecture Notes - Lecture 30: Scilab
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Math 1b: calculus - lecture 30: finishing up variation of parameters (r-3)(r+2) = 0. Q: solve (d2y/dx2) - (dy/dx) - 6y = 1 + e-2x. A: we have (d2y/dx2) - (dy/dx) - 6y = 1 + e-2x. The corresponding homogeneous equation is (d2y/dx2) - (dy/dx) - 6y = 0. This has auxiliary equation r2 - r - 6 = 0. So the auxiliary equation has solutions r=3 and r=-2. So the homogeneous equation has general solution y = ae3x + be-2x. To find a particular solution, let"s guess y = c + dxe-2x. So (dy/dx) = -2dxe-2x + de-2x. (d 2y/dx2) = -2de-2x - 2de-2x + 4dxe-2x. So (d2y/dx2) - (dy/dx) - 6y = (-4d - d)e-2x + (4d+2d-6d)xe-2x + c. For this to give a solution to [*], we require -5d = 1 and -6c = 1. So the general solution to * is y = ae3x + be-2x -
