I have to find the particular solution to the differential equation: (-21/4)y''+2y'+y=4xe^(3x) First, I chose my trial function to be yp=(Ax+B)*e^(3x). Is this correct??? so yp'=3(Ax+B)*e^(3x) yp''=9(Ax+B)*e^(3x) So I plug these into the differential equation and I get: (-189/4)Axe^(3x)-(189/4)Be^(3x)+6Axe^(3x)+6Be^(3x)+Axe^(3x)+Be^(3x)=4xe^(3x) I group like terms and I get A=-16/161 and B=0 So yp=(-16/161)Axe^(3x) This is not correct. Can someone please tell me where I'm going wrong? Thanks!