This question gives a tidier way of using the method of undetermined coefficients for polynomials times...

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This question gives a tidier way of using the method of undetermined coefficients for polynomials times exponentials. Write p(r) for the polynomial p(r) = r² + br+c. For a function , define: d² d L[0] =2+b+co dt = L is a linear differential operator. The kernel of L, ker L { L[] = 0} is equal to the set of solutions to the ODE y" + by' + cy = 0. (see the beginning of §3.2 in the textbook, page 110, for more details) (a) Suppose that g(t) is an arbitrary polynomial in t. Show that: L[g(t)ert] = (q"(t) + p'(r)q'(t) + p(r)q(t)) ert (Note: the derivatives on q are taken with respect to t, and the derivatives on p are taken with respect to r, so that p'(r) = 2r + b). (b) Use the result of the previous part to find a solution to: y" + 2y' 15y = (-7t² - 9t+ 6)e²t (Hint: identify p(r), then use the previous part to get an equation for q(t) from L[q(t)et] = (t²-t + 1)et. Write q(t) = at² + bt + c and solve for a, b, c). (c) Use the result of part a again to find a solution to: 3t y' + 2y' 15y = (4t² - 15t + 2)e³t (Hint: this is much the same as part b, but notice that now 3 is a root of p(r)). This question gives a tidier way of using the method of undetermined coefficients for polynomials times exponentials. Write p(r) for the polynomial p(r) = r² + br+c. For a function , define: d² d L[0] =2+b+co dt = L is a linear differential operator. The kernel of L, ker L { L[] = 0} is equal to the set of solutions to the ODE y" + by' + cy = 0. (see the beginning of §3.2 in the textbook, page 110, for more details) (a) Suppose that g(t) is an arbitrary polynomial in t. Show that: L[g(t)ert] = (q"(t) + p'(r)q'(t) + p(r)q(t)) ert (Note: the derivatives on q are taken with respect to t, and the derivatives on p are taken with respect to r, so that p'(r) = 2r + b). (b) Use the result of the previous part to find a solution to: y" + 2y' 15y = (-7t² - 9t+ 6)e²t (Hint: identify p(r), then use the previous part to get an equation for q(t) from L[q(t)et] = (t²-t + 1)et. Write q(t) = at² + bt + c and solve for a, b, c). (c) Use the result of part a again to find a solution to: 3t y' + 2y' 15y = (4t² - 15t + 2)e³t (Hint: this is much the same as part b, but notice that now 3 is a root of p(r)).