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Question Number 207991 by efronzo1 last updated on 02/Jun/24

$$\mathrm{y}\underset{―}{\underbrace{\mathcal{W}}}$$

Answered by mr W last updated on 02/Jun/24

$$particularsolution:$$$$y=A\sin x+B\cos x$$$$y^{\prime }=A\cos x-B\sin x$$$$y^{″}=-A\sin x-B\cos x$$$$-3A-B=1$$$$-A-3B=0$$$$A=\frac{3}{10}$$$$B=-\frac{1}{10}$$$$\Rightarrow y=\frac{1}{10}(3\sin x-\cos x)$$$$generalsolution:$$$$y^{″}+y^{\prime }-2y=0$$$$r^2+r-2=0$$$$(r+2)(r-1)=0$$$$r=-2,1$$$$y=C_1{e}^{-2x}+C_2{e}^x$$$$$$$$\Rightarrow y=C_1{e}^{-2x}+C_2{e}^x+\frac{1}{10}(3\sin x-\cos x)$$

Answered by efronzo1 last updated on 02/Jun/24

$$({\mathrm{D}}^2+\mathrm{D}-2)\mathrm{y}=\sin \mathrm{x}$$$$\mathrm{Let}{\mathrm{D}}^2+\mathrm{D}-2=0$$$$(\mathrm{D}+2)(\mathrm{D}-1)=0$$$$\mathrm{D}=-2,\mathrm{D}=1$$$${\mathrm{y}}_{\mathrm{h}}={\mathrm{C}}_1{\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{C}}_2{\mathrm{e}}^{\mathrm{x}}$$$$$$$${\mathrm{y}}_{\mathrm{p}}=\frac{1}{{\mathrm{D}}^2+\mathrm{D}-1}\left(\sin \mathrm{x}\right)$$$${\mathrm{y}}_{\mathrm{p}}=\frac{1}{\mathrm{D}-3}\left(\sin \mathrm{x}\right)$$$${\mathrm{y}}_{\mathrm{p}}=\frac{\mathrm{D}+3}{{\mathrm{D}}^2-9}\left(\sin \mathrm{x}\right)=-\frac{\mathrm{D}+3}{10}\left(\sin \mathrm{x}\right)$$$${\mathrm{y}}_{\mathrm{p}}=-\frac{1}{10}(\cos \mathrm{x}+3\sin \mathrm{x})$$$${\mathrm{y}}_{\mathrm{g}}={\mathrm{C}}_1{\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{C}}_2{\mathrm{e}}^{\mathrm{x}}-\frac{1}{10}(\cos \mathrm{x}+3\sin \mathrm{x})$$

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