Question

Using the method of undetermined coefficients find the P.I for the D.E x’’’(t) – x’’(t) = 3e^t + sin⁡t.

(a) xp = 3e^t +  \(\frac{t}{2}\)  (cos⁡t – 2 sin⁡t )

(b) xp = 3te^t +  \(\frac{1}{2}\)  (cos⁡t + sin⁡t )

(c) xp = 3te^t +  \(\frac{t}{3}\) (4cos⁡t + 2sin⁡t )

(d) xp = 3e^t +  \(\frac{1}{2}\) (cos⁡t – sin⁡t )

The question was posed to me in semester exam.

I need to ask this question from Method of Undetermined Coefficients in portion Linear Differential Equations – Second and Higher Order of Engineering Mathematics

Answer 1

Right answer is (b) xp = 3te^t +  \(\frac{1}{2}\)  (cos⁡t + sin⁡t )

To explain I would say: We have (D^3 – D^2)x(t) =  3e^t + sin⁡t, where D = d/dt, A.E is m^3 – m^2 = 0

                        m^2 (m – 1) = 0 –> m = 0, 0, 1 –> xc (t) = (c1 + c2 t) + c3 e^t

∅(t) =  3e^t + sin⁡t  we assume for P.I in the form xp = ate^t + b cos⁡t + c sin⁡t …(1)

since 1 is a root and ∓i are not a roots of the A.E. To find a, b& c such that

xp’’’(t) –   xp’’(t) = 3e^t + sin⁡t……(2)

from (1) we have xp‘ =  a(te^t + e^t) – b sin⁡t + c cos⁡t

xp” =   a(te^t + 2e^t) – b cos⁡t – c sin⁡t

xp”’ =  a(te^t + 3e^t) + b sin⁡t – c cos⁡t,now (2) becomes

ate^t + 3ae^t + b sin⁡t – c cos⁡t, –  ate^t – 2ae^t + b cos⁡t + c sin⁡t =  3e^t + sin⁡t

 ae^t + (b + c)  sin⁡t + (b – c)  cos⁡t =  3e^t + sin⁡t

 – –  > a = 3 and b + c = 1, b – c = 0 –> a = 3 and b = 1/2, c = 1/2

hence from (1) xp = 3te^t + 1/2  (cos⁡t + sin⁡t) is the particular integral solution.