At first, we solve the homogeneous equation: g′′−g′+2g=0. The characteristic equation has the form: λ2−λ+2=0. The solution is: λ1=21−i7, λ2=21+i7. Thus, g=C1eλ1x+C2eλ2xwith C1,C2∈C is the solution of equation: g′′−g′+2g=0. The solution can be rewritten in the following form: g=k1e21xsin(27x)+k2e21xcos(27x) with k1,k2∈C. We will try to find a solution of inhomogeneous equation in the form: f=aeαx+beβx with a,b,α,β∈C . We receive: f′=aαeαx+bβeβx, f′′=aα2eαx+bβ2eβx. f′′−f′+2f=a(α2−α+2)eαx+b(β2−β+2)eβx . Right side of equation has the form: e2x+e−2x. We put α=2, β=−2 and receive: f′′−f′+2f=4ae2x+8be−2x=e2x+e−2x. From the latter we get: a=41, b=81. A solution is f=41e2x+81e−2x. The general solution of the initial equation is: y=f+g=k1e21xsin(27x)+k2e21xcos(27x)+41e2x+81e−2x with k1,k2∈C.
Answer: the solution is: y=k1e21xsin(27x)+k2e21xcos(27x)+41e2x+81e−2x with k1,k2∈C.
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