y′′−4xy′+(4x2−1)y=−3ex2sin(2x) consider:y′′−4xy′+(4x2−1)y=0y(x)=ex2⋅z(x)(2ex2z+4x2ex2z+4xex2z′+ex2z′′)−4x(2xex2z+ex2z′)+(4x2−1)ex2z=0z′′+z=0z=c1cos(x)+c2sin(x)y(x)=c1ex2cos(x)+c2ex2sin(x)
check the particular solution:y1=ex2sin(2x)y′=2xex2sin(2x)+2ex2cos(2x)y′′=−2ex2sin(2x)+4x2ex2sin(2x)+8xex2cos(2x)y′′−4xy′+(4x2−1)y=−2ex2sin(2x)+4x2ex2sin(2x)+8xex2cos(2x)−4x(2xex2sin(2x)+2ex2cos(2x))+(4x2−1)(ex2sin(2x))=−3ex2sin(2x)
then solution is:y(x)=c1ex2cos(x)+c2ex2sin(x)+ex2sin(2x)
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