By separating variables and integrating,
#int e^{-y}cosy dy=int (1+x^2)e^{-x} dx#
Integration by Parts 1
#u=e^{-y}" "dv=cosy dy#.
#du=-e^{-y} dy" "v=siny#.
Integration by Parts 2
#u=e^{-y}" "dv=siny dy#.
#du=-e^{-y}dy" "v=-cosy#
By Integration by Pats 1,
#(LHS)=e^{-y}siny+int e^{-y}siny dy#
by Integration by Parts 2,
#=e^{-y}siny-e^{-y}cosy-int e^{-y}cosy dy#
Since the last integral is the same as #(LHS)#, we have
#(LHS)=e^{-y}(siny-cosy)-(LHS)#
by adding #(LHS)#,
#Rightarrow2(LHS)=e^{-y}(siny-cosy)#
by dividing by #2#,
#Rightarrow (LHS)=e^{-y}/2(siny-cosy)#
Integration by Parts 3
#u=1+x^2" "dv=e^{-x}dx#
#du=2xdx" "v=-e^{-x}#
Integration by Parts 4
#u=2x" "dv=e^{-x}dx#
#du=2dx" "v=-e^{-x}#
Let us evaluate the right-hand side.
By Integration by Parts 3,
#(RHS)=-(1+x^2)e^{-x}+int2xe^{-x}dx#
by Integration by Parts 4,
#=-(1+x^2)-2xe^{-x}+int 2e^{-x} dx#
#=-(x^2+2x+1)e^{-x}-2e^{-x}+C#
#=-(x^2+2x+3)e^{-x}+C#
By setting #(LHS)=(RHS)#,
#e^{-y}/2(siny-cosy)=-(x^2+2x+3)e^{-x}+C#
Since #y(0)=0#, we have
#Rightarrow 1/2(0-1)=-3+C Rightarrow C=5/2#
Hence, the solution is implicitly expressed as
#e^{-y}/2(siny-cosy)=-(x^2+2x+3)e^{-x}+5/2#.
I hope that this was helpful.
