Question

how to solve y''+ y = 2cosx​

Answer 1

Answer:

yy=y¯¯¯+y⋆=C1cos(x)+C2sin(x)+x(Acos(x)+Bsin(x))(8)

Left=y′′+y=−C1cos(x)–––––––––−C2sin(x)–––––––––−Asin(x)+Bcos(x)−Axcos(x)–––––––––−Asin(x)+Bcos(x)−Bxsin(x)––––––––––+C1cos(x)–––––––––+C2sin(x)–––––––––+Axcos(x)–––––––––+Bxsin(x)––––––––––=−2Asin(x)+2Bcos(x)=2cos(x)=Right(9)

Thus A and B can be found out, which are A=0,B=1, repectively.

Conclusion, the general solution of this differential equation is

y=C1cos(x)+C2sin(x)+xsin(x)(10)