This question will show an efficient way for using the method of undetermined coefficients when the non-homogeneous part is of the form eat cos(bt) or eat sin(bt) (and this also works if a = 0). Combined with the methods of the previous question this is also useful when the non-homogeneous part is a polynomial times a function of the form just mentioned. Consider the ODE: y" - y' - 6y = e cos(t) (a) Use the method of undetermined coefficients to find a solution of y"-y'-6y = e(+i)t (Hint: this can be done the exact same way as the usual method of undetermined coefficients, but will involve complex numbers). (b) With o(t) a solution to y" - y' - 6y = e(+i)t, calculate Reo(t). (Hint: remember to use Euler's formula to expand the exponential. When trying to take the real or imaginary part of a product of complex numbers it helps to expand it out and collect all the real and imaginary terms together) (c) Explain why Re o(t) is a solution of y" - y' - 6y = et cos(t). (d) Instead use the imaginary part to find a solution to y" - y' - 6y = et sin(t) Hints: Remember that for a complex number z = a +ib, the real and imaginary parts are defined by: Reza, Im z = b. If : RC is a function, then o(t) = u(t) +iv(t) with u, v : R R two real valued functions, and Reo(t) = u(t), Im o(t) = v(t). If : R C is a function, then with u, v as in the previous point, then is differentiable if and only if u and u are both differentiable, and '(t) = u'(t) + iv' (t). 1 a + ib a - ib a +6