Community Answer

Given D*E is (2x+1)y^('')+2(4x+1)y^(')+8xy=0rarr(1)To show y_(1)=e^(-2x) is a solution of equ (1).Now y_(1)^(')=-2e^(-2x)quad&quady_(1)^('')=-2e^(-2x)(-2)=4e^(-2x)we plugging these in (1), we get{:[(2x+1)4e^(-2x)+2(4x+1)(-2e^(-2x))+8xe^(-2x)=0],[=>e^(-2x)(8x+4-16 x-4+8x)=0],[=>e^(-2x)(0)=0],[=>0=0]:}:.y_(1)=e^(-2x) is a solution to given D*EFroon (1), we haney^('')+(2(4x+1))/(2x+1)y^(')+(8x)/(2x+1)y=0compare y^('')+p(x)y^(')+Q(x)y=0, we havep(x)=(2(4x+1))/(2x+1)quad&quad Q(x)=(8x)/(2x+1)By the reduction of order, we hane.{:[y_(2)=y_(1)int(e^(-int p(x)dx))/(y_(1)^(2))dx],[=e^(-2x)int(e^(-int(2(4x+1))/(2x+1))dx)/((e^(-2x))^(2))dx]:}{:[" 2x+1) "{:[4x+1(2],[(4x+2)/((-1))]:}],[=e^(-2x)int(e^(-2(2-(1)/(2x+1))))/(e^(-4x))dx],[=e^(-2x)int(e^(-2(2x-(1)/(2)log(2x+1))))/(e^(-4x))dx],[=e^(-2x)int(e^( ... See the full answer