MATH 256 Chapter Notes - Chapter 3: Polynomial, Scilab, Autocad Dxf
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Recall the ode: with constants a, b and c, and f (x) a prescribed function. ay00 + by0 + cy = f (x) Notice that if y / ex, then all three terms in the ode have a ex that therefore can be cancelled out. i. e. a potential solution to the ode is y(x) = dex , for some constant d. 2 ex is a solution to the ode. Indeed, plugging and chugging implies d = 1 general solution we are expecting some additional bits with arbitrary constants in them. Now let y(x) = yh(x) + yp(x) with yp(x) = 1. 2 ex and yh(x) satisfying the corresponding homogeneous. On plugging this combo into the ode, one realizes that it will still satisfy the. Ode, yh equation (because the yh(x) cancels on the lhs, and yp(x) gives us the rhs). But we know how to solve for yh(x), and it, in general, will contain two arbitrary constants!
