1.The given equation is "(D^{2}-2D+5)y = e^{x}\\sin x".
The auxiliary equation is "m^{2}-2m+5=0." Solving for m, we get "m=1\\pm 2i".
Thus, the complementary function is "C.F = e^{x}(c_{1}\\cos 2x + c_{2}\\sin 2x)."
To find the particular integral, let "y_{p} = Ae^{x}\\sin x" be the trial solution.
Then,
"y'_{p} = Ae^{x}(\\cos x + \\sin x)~\\text{and}~\\\\ \ny''_{p} = A\\bigg(e^{x}(-\\sin x+\\cos x )+e^{x}(\\cos x + \\sin x)\\bigg)\\\\ ~~~~~= 2Ae^{x}\\cos x"
Therefore,
"2Ae^{x}\\cos x - 2\\bigg(Ae^{x}(\\cos x+\\sin x)\\bigg)+5Ae^{x}\\sin x = e^{x}\\sin x\\\\\nA = \\dfrac{1}{3}"
Hence the particular integral is "y_{p} = \\dfrac{1}{3}e^{x}\\sin x" .
The general solution is "y=e^{x}(c_{1}\\cos 2x+c_{2}\\sin 2x)+\\dfrac{1}{3}e^{x}\\sin x"
2.The given equation is "(D^{2}-D-2)y=e^{2x}."
The auxiliary equation is "m^{2}-m-2=0". Solving for m, we get "m=-1,2".
Thus, the complementary function is "C.F=c_{1}e^{-x}+c_{2}e^{2x}".
To find the particular integral, since "g(x) =e^{2x}" is a linear combination of a term in complementary function we choose "y_{p} = Axe^{2x}" as the trial particular solution.
Thus,
"y'_{p}=A(2xe^{x}+e^{2x})=Ae^{2x}(2x+1)\\\\\ny''_{p}=A\\bigg(2(2xe^{x}+e^{2x})+2e^{2x}\\bigg)=4Ae^{2x}(x+1)\\\\"
Therefore,
"4Ae^{2x}(x+1)-Ae^{2x}(2x+1)-2Axe^{2x}=e^{2x}\\\\\nA(4x+4-2x-1-2x)=1\\\\\nA=\\dfrac{1}{3}"
Hence the particular integral is, "y_{p}=\\dfrac{1}{3}xe^{2x}"
The general solution is, "y=c_{1}e^{-x}+e^{2x}(\\frac{x}{3}+c_{2})"
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