MATH 1B Lecture 28: Challenging 2nd Order Nonhomogeneous Differential Equations
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Math 1b: calculus - lecture 28: challenging 2nd order nonhomogeneous equations r = 2i, -2i. A: we have (d2y/dx2) + 4y = sin(2t). The corresponding homogeneous equation is: (d2y/dx2) + 4y = 0. This has an auxiliary equation r2 + 4 = 0. So, the auxiliary equation has 2 solutions, r = -2i and r = 2i. So, the homogeneous equation has a general solution y = acos(2t) + bsin(2t), where a and b are constants. To find a particular solution to *, let"s guess y = t(acos(2t) + bsin(2t)). So (dy/dx) = acos(2t) + bsin(2t) + t(-2asin(2t) + 2bcos(2t)). So (d2y/dx2) = -2asin(2t) + 2bsin(2t) + (-2asin(2t) + 2bcos(2t)) + t(-4sin(2t) - 4bsin(2t)). We want this to equal sin(2t), so we may take a = - and b = 0. So a particular solution to * is y = (-tcos(2t))/4. So the general solution to * is y = acos(2t) + bsin(2t) - [(-tcos(2t))/4].
