Solve one of the following two non-homogeneous differential equations using whatever technique your prefer. Put an "X" through the equation you would not like me to grade. If you do not technique you prefer. Put an "X" through the equation you would not like me to grade. If you do not put an "X" through one of the equations, I will grade whichever problem I prefer to grade. a) y" - 4y + 4y = 6xe^2x b) y" + 9y = 5 cos x - 7 sin x

Answer:a.[tex]y(x)=c_1e^{2x}+c-2xe^{2x}+x^3e^{2x}[/tex]b[tex]y(x)=c_1cos 3x+c_2 sin 3x-5 cos x+ 7sin x[/tex]Step-by-step explanation:1.[tex]y''-4y'+4y=6x e^{2x}[/tex]Auxillary equation [tex]D^2-4D+ 4=0[/tex][tex](D-2)(D-2)=0[/tex]D=2,2Then complementary solution =[tex] C_1e^{2x}+C_2xe^{2x}[/tex]Particular solution [tex]=\frac{6 xe^{2x}}{(D-2)^2}[/tex]D is replace by D+2 then we get P.I=[tex]\frac{6xe^{2x}}{0}[/tex]P.I=[tex]\frac{e^{ax}}{D+a} \cdot .V[/tex]where V is a function of xP.I=[tex]\frac}x^3e^{2x}[/tex]By integrating two times Hence, the general solution [tex]y(x)=c_1e^{2x}+c-2xe^{2x}+x^3e^{2x}[/tex]b.y''+9y=5 cos x-7 sin xAuxillary equation [tex]D^2+9=0[/tex]D=[tex]\pm 3i[/tex][tex]C.F=c_1 cos 3x+ c_2sin 3x[/tex]P.I=[tex]\frac{5 cos x-7 sin x}{D^2+9}[/tex]P.I=[tex]\frac{sin ax}{D^2+bD +C}[/tex]Then  D square is replace by -a square [tex] D^2 [/tex] is replace by - then we getP.I=-5 cos x+7 sin xThe general solution[tex]y(x)=c_1cos 3x+c_2 sin 3x-5 cos x+ 7sin x[/tex]