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Question Number 119233 by bemath last updated on 23/Oct/20

$$(D^2-5D+6)y=e^{3x}$$

Answered by benjo_mathlover last updated on 23/Oct/20

$$(D-2I)(D-3I)\left(y\right)=e^{3x};I\left(y\right)=y$$$$let(D-3I)\left(y\right)=z$$$$\Rightarrow (D-2I)\left(z\right)=e^{3x}$$$$\Rightarrow z^{\prime }-2z=e^{3x};\left(e^{-2x}\right)(z^{\prime }-2z)=e^x$$$$\Rightarrow e^{-2x}.z^{\prime }-2e^{-2x}.z=e^x$$$$\Rightarrow {(e^{-2x}.z)}^{\prime }=e^x$$$$\Rightarrow e^{-2x}.z=\int e^{x}dx$$$$\Rightarrow e^{-2x}.z=e^x+A;z=e^{3x}+{Ae}^{2x}$$$$\Rightarrow (D-3I)y=e^{3x}+{Ae}^{2x}$$$$\Rightarrow y^{\prime }-3y=e^{3x}+{Ae}^{2x}$$$$\Rightarrow \left(e^{-3x}\right)(y^{\prime }-3y)=1+{Ae}^{-x}$$$$\Rightarrow {(e^{-3x}.y)}^{\prime }=1+{Ae}^{-x}$$$$\Rightarrow e^{-3x}.y=x-{Ae}^{-x}+B$$$$\Rightarrow y={xe}^{3x}-{Ae}^{2x}+{Be}^{3x}.$$$$$$

Commented by bemath last updated on 23/Oct/20

$$great$$

Answered by TANMAY PANACEA last updated on 23/Oct/20

$$C.F$$$$y=e^{mx}$$$$m^2-5m+6=0$$$$(m-2)(m-3)=0\to m=2,3$$$$C.F={Ae}^{2x}+{Be}^{3x}$$$$P.I$$$$y=\frac{e^{3x}}{(D-3)(D-2)}$$$$\mathit{y}=\frac{e^{3x}}{(3-2)}\times \frac{1}{(D+3-3)}=\frac{e^{3x}}{1}\times x$$$$=\frac{e^{3x}}{1}\times x$$$$completesolution$$$$y={Ae}^{2x}+{Be}^{3x}+\frac{{xe}^{3x}}{1}$$$$$$$$$$

Answered by mathmax by abdo last updated on 24/Oct/20

$${\mathrm{y}}^{{}^{″}}-{5\mathrm{y}}^{{}^{\prime }}+6\mathrm{y}={\mathrm{e}}^{3\mathrm{x}}$$$$\mathrm{h}\to {\mathrm{r}}^2-5\mathrm{r}+6=0\to \mathrm{\Delta }=25-24=1\Rightarrow {\mathrm{r}}_1=\frac{5+1}{2}=3$$$${\mathrm{r}}_2=\frac{5-3}{2}=1\Rightarrow {\mathrm{y}}_{\mathrm{h}}={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{3\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{3\mathrm{x}}\\ {\mathrm{e}}^{\mathrm{x}}{3\mathrm{e}}^{3\mathrm{x}}\end{array}\right|={3\mathrm{e}}^{4\mathrm{x}}-{\mathrm{e}}^{4\mathrm{x}}={2\mathrm{e}}^{4\mathrm{x}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}0{\mathrm{e}}^{3\mathrm{x}}\\ {\mathrm{e}}^{3\mathrm{x}}{3\mathrm{e}}^{3\mathrm{x}}\end{array}\right|=-{\mathrm{e}}^{6\mathrm{x}}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}0\\ {\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{3\mathrm{x}}\end{array}\right|={\mathrm{e}}^{4\mathrm{x}}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{W}}_1}{\mathrm{W}}\mathrm{dx}=\int \frac{-{\mathrm{e}}^{6\mathrm{x}}}{{2\mathrm{e}}^{4\mathrm{x}}}\mathrm{dx}=-\frac{1}{2}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}=-\frac{1}{4}{\mathrm{e}}^{2\mathrm{x}}$$$${\mathrm{v}}_2=\int \frac{{\mathrm{W}}_2}{\mathrm{W}}\mathrm{dx}=\int \frac{{\mathrm{e}}^{4\mathrm{x}}}{{2\mathrm{e}}^{4\mathrm{x}}}\mathrm{dx}=\frac{\mathrm{x}}{2}\Rightarrow$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2={\mathrm{e}}^{\mathrm{x}}(-\frac{1}{4}{\mathrm{e}}^{2\mathrm{x}})+{\mathrm{e}}^{3\mathrm{x}}\left(\frac{\mathrm{x}}{2}\right)=-\frac{1}{4}{\mathrm{e}}^{3\mathrm{x}}+\frac{\mathrm{x}}{2}{\mathrm{e}}^{3\mathrm{x}}\Rightarrow$$$$\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}\mathrm{y}={\mathrm{y}}_{\mathrm{p}}+{\mathrm{y}}_{\mathrm{h}}$$$$\Rightarrow \mathrm{y}=(\frac{\mathrm{x}}{2}-\frac{1}{4}){\mathrm{e}}^{3\mathrm{x}}+{\mathrm{ae}}^{\mathrm{x}}+\mathrm{b}{\mathrm{e}}^{3\mathrm{x}}$$

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