Generalized Exponential Response Formula We can also solve an LTI DE p( D ) x = q(t) with exponential input q(t) = Be at even when p( a) = 0. The answer is given by the following generalized Exponential Response formula (the proof of which we post pone to the session on Linear Operators. Generalized Exponential Response Formula. Let p( D ) be a polynomial operator with constant coefﬁcients, and p(s) its s-th derivative. Then p( D ) x = Be at , has the particular solution ⎧ Be at ⎪ ⎪ ⎪ ⎪ ⎪ p( a) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Bte at ⎪ ⎪ ⎪ ⎪ � ⎪ ⎨ p ( a) xp = Bt2 e at ⎪ ⎪ ⎪ ⎪ ⎪ p�� ( a) ⎪ ⎪ ⎪ ... ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Bts e at ⎪ ⎪ ⎩ (s) p ( a) where a is real or complex if p( a) �= 0 if p( a) = 0 and p� ( a) �= 0 if p( a) = p� ( a) = 0 and p�� ( a) �= 0 if a is an s-fold zero Note: Later when we cover resonance the case p( a) = 0, p� ( a) �= 0 will be called the Resonant Response Formula Example 1. Find a particular solution to the equation .. . x + 8x + 15x = e−5t Solution. The characteristic polynomial is p(r ) = r2 + 8r + 15. Since p(−5) = 0 we need to use the generalized ERF. Computing p� (r ) = 2r + 8, which implies p� (−5) = −2. Therefore the gen eralized ERF gives te−5t te−5t xp = � =− . p (−5) 2 Example 2. Find a particular solution to .. . x + 2x + 2x = e−t cos t. Exponential Response Formula 18.03SC Solution. First we complexify the equation .. . z + 2z + 2z = e(−1+i)t , where x = Re(z). The characteristic polynomial is p(r ) = r2 + 2r + 2. Computing, p(−1 + i ) = (−1 + i )2 + 2(−1 + i ) + 2 = 0, p� (r ) = 2r + 2, Since p(−1 + i ) = 0 we use the generalized ERF zp = te(−1+i)t te(−1+i)t te−t (cos t + i sin t) = = p� (−1 + i ) 2i 2i Finally we take the real part to get x p = Re(z p ) = te−t sin t . 2 p� (−1 + i ) = 2i.