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Question Number 91911 by jagoll last updated on 03/May/20

$${\mathrm{y}}^{⁗}+{2\mathrm{y}}^{″}+\mathrm{y}=\sin \mathrm{x}$$

Answered by niroj last updated on 03/May/20

$${\mathrm{y}}^{{}^{⁗}}+{2\mathrm{y}}^{{}^{″}}+\mathrm{y}=\sin \mathrm{x}$$$$\frac{{\mathrm{d}}^4\mathrm{y}}{{\mathrm{dx}}^4}+2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\mathrm{y}=\sin \mathrm{x}$$$$({\mathrm{D}}^4+{2\mathrm{D}}^2+1)\mathrm{y}=\sin \mathrm{x}$$$$\mathrm{A}.\mathrm{E}.,{\mathrm{m}}^4+{2\mathrm{m}}^2+1=0$$$${({\mathrm{m}}^2+1)}^2=0$$$$\mathrm{m}=0\stackrel{-}{+}\mathrm{i},0\stackrel{-}{+}\mathrm{i}$$$$\mathrm{CF}=({\mathrm{C}}_1\mathrm{x}+{\mathrm{C}}_2)\cos \mathrm{x}+({\mathrm{C}}_3\mathrm{x}+{\mathrm{C}}_4)\mathrm{sinx}$$$$\mathrm{PI}=\frac{\sin \mathrm{x}}{{\mathrm{D}}^4+{2\mathrm{D}}^2+1}=\frac{\sin \mathrm{x}}{{({\mathrm{D}}^2+1)}^2}$$$$=\frac{\mathrm{x}\sin \mathrm{x}}{{4\mathrm{D}}^3+4\mathrm{D}}=\frac{\mathrm{x}\sin \mathrm{x}}{2({\mathrm{D}}^2+1)2\mathrm{D}}$$$$=\frac{{\mathrm{x}}^2\mathrm{sinx}}{{12\mathrm{D}}^2+4}=\frac{{\mathrm{x}}^2\mathrm{sinx}}{4({3\mathrm{D}}^2+1)}$$$$=\frac{({3\mathrm{D}}^2-1){\mathrm{x}}^2\sin \mathrm{x}}{4({9\mathrm{D}}^2-1)}$$$$=\frac{({3\mathrm{D}}^2-1){\mathrm{x}}^2\sin \mathrm{x}}{4(-9-1)}=-\frac{1}{40}({3\mathrm{D}}^2-1){\mathrm{x}}^2\mathrm{sinx}$$$$-\frac{1}{40}(6\sin \mathrm{x}+12\mathrm{x}\cos \mathrm{x}-{3\mathrm{x}}^2\sin \mathrm{x}-{\mathrm{x}}^2\sin \mathrm{x})$$$$-\frac{1}{40}(6\sin \mathrm{x}+12\mathrm{x}\cos \mathrm{x}-{4\mathrm{x}}^2\mathrm{sinx})$$$$\mathrm{y}=({\mathrm{C}}_1\mathrm{x}+{\mathrm{C}}_2)\cos \mathrm{x}+({\mathrm{C}}_2\mathrm{x}+{\mathrm{C}}_4)\sin \mathrm{x}-\frac{1}{40}(6\sin \mathrm{x}+12\mathrm{x}\cos \mathrm{x}-{4\mathrm{x}}^2\sin \mathrm{x})$$

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