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Question Number 134848 by mohammad17 last updated on 07/Mar/21

Commented by mohammad17 last updated on 07/Mar/21

$$solvebylaplasetransformsir$$

Commented by mohammad17 last updated on 07/Mar/21

$$helpmesir$$

Answered by mathmax by abdo last updated on 07/Mar/21

$$\mathrm{e}\Rightarrow \mathrm{L}\left({\mathrm{y}}^{{}^{″}}\right)-3\mathrm{L}\left({\mathrm{y}}^{{}^{\prime }}\right)+2\mathrm{L}\left(\mathrm{y}\right)=4\mathrm{L}\left({\mathrm{e}}^{2\mathrm{x}}\right)\Rightarrow$$$${\mathrm{x}}^2\mathrm{L}\left(\mathrm{y}\right)-{\mathrm{xy}}^{{}^{\prime }}\left(0\right)-{\mathrm{y}}^{{}^{″}}\left(0\right)-3(\mathrm{xL}\left(\mathrm{y}\right)-{\mathrm{y}}^{{}^{\prime }}\left(0\right))+2\mathrm{L}\left(\mathrm{y}\right)=4\mathrm{L}\left({\mathrm{e}}^{2\mathrm{x}}\right)\Rightarrow$$$$({\mathrm{x}}^2-3\mathrm{x}+2)\mathrm{L}\left(\mathrm{y}\right)={\mathrm{xy}}^{{}^{\prime }}\left(0\right)+{\mathrm{y}}^{{}^{″}}\left(0\right)-{3\mathrm{y}}^{{}^{\prime }}\left(0\right)+4\mathrm{L}\left({\mathrm{e}}^{2\mathrm{x}}\right)\Rightarrow$$$$({\mathrm{x}}^2-3\mathrm{x}+2)\mathrm{L}\left(\mathrm{y}\right)=5\mathrm{x}-15+{\mathrm{y}}^{{}^{″}}\left(0\right)+4\mathrm{L}\left({\mathrm{e}}^{2\mathrm{x}}\right)$$$$\mathrm{we}\mathrm{have}\mathrm{L}\left({\mathrm{e}}^{2\mathrm{x}}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{2\mathrm{t}}{\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(2-\mathrm{x})\mathrm{t}}\mathrm{dt}={\left[\frac{1}{2-\mathrm{x}}{\mathrm{e}}^{(2-\mathrm{x})\mathrm{t}}\right]}_{\mathrm{t}=0}^{\mathrm{\infty }}$$$$=-\frac{1}{2-\mathrm{x}}\Rightarrow ({\mathrm{x}}^2-3\mathrm{x}+2)\mathrm{L}\left(\mathrm{y}\right)=\frac{4}{\mathrm{x}-2}+5\mathrm{x}-15+{\mathrm{y}}^{{}^{″}}\left(0\right)\Rightarrow$$$$\mathrm{L}\left(\mathrm{y}\right)=\frac{4}{(\mathrm{x}-2)({\mathrm{x}}^2-3\mathrm{x}+2)}+\frac{5\mathrm{x}-15}{{\mathrm{x}}^2-3\mathrm{x}+2}+\frac{{\mathrm{y}}^{\left(2\right)}\left(0\right)}{{\mathrm{x}}^2-3\mathrm{x}+2}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)={4\mathrm{L}}^{-1}\left(\frac{1}{(\mathrm{x}-2)({\mathrm{x}}^2-3\mathrm{x}+2)}\right)+{\mathrm{L}}^{-1}\left(\frac{5\mathrm{x}-15}{{\mathrm{x}}^2-3\mathrm{x}+2}\right)+{\mathrm{y}}^{\left(2\right)}\left(0\right){\mathrm{L}}^{-1}\left(\frac{1}{{\mathrm{x}}^2-3\mathrm{x}+2}\right)$$$${\mathrm{x}}^2-3\mathrm{x}+2=0\to \mathrm{\Delta }=1\Rightarrow {\mathrm{x}}_1=\frac{3+1}{2}=2\mathrm{and}{\mathrm{x}}_2=\frac{3-1}{2}=1\Rightarrow$$$$\frac{1}{{\mathrm{x}}^2-3\mathrm{x}+2}=\frac{1}{(\mathrm{x}-2)(\mathrm{x}+1)}=\frac{1}{3}(\frac{1}{\mathrm{x}-2}-\frac{1}{\mathrm{x}+1})\Rightarrow$$$${\mathrm{L}}^{-1}\left(\frac{1}{{\mathrm{x}}^2-3\mathrm{x}+2}\right)=\frac{1}{3}{\mathrm{e}}^{2\mathrm{x}}-\frac{1}{3}{\mathrm{e}}^{-\mathrm{x}}\mathrm{also}$$$$\frac{5\mathrm{x}-15}{{\mathrm{x}}^2-3\mathrm{x}+2}=\frac{\mathrm{a}}{\mathrm{x}-2}+\frac{\mathrm{b}}{\mathrm{x}+1}\Rightarrow {\mathrm{L}}^{-1}(....)={\mathrm{ae}}^{2\mathrm{x}}+{\mathrm{be}}^{-\mathrm{x}}$$$$.....\mathrm{be}\mathrm{continued}....$$$$$$

Commented by mathmax by abdo last updated on 07/Mar/21

$$\mathrm{sorry}\frac{1}{{\mathrm{x}}^2-3\mathrm{x}+2}=\frac{1}{(\mathrm{x}-2)(\mathrm{x}-1)}=\frac{1}{\mathrm{x}-2}-\frac{1}{\mathrm{x}-1}\Rightarrow$$$${\mathrm{L}}^{-1}\left(\frac{1}{{\mathrm{x}}^2-3\mathrm{x}+2}\right)={\mathrm{e}}^{2\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}$$

Answered by mathmax by abdo last updated on 07/Mar/21

$$\mathrm{wronskien}\mathrm{method}$$$$\mathrm{h}\to {\mathrm{y}}^{\left(2\right)}-{3\mathrm{y}}^{\left(1\right)}+2\mathrm{y}=0\to {\mathrm{r}}^2-3\mathrm{r}+2=0\Rightarrow {\mathrm{r}}_1=1\mathrm{and}{\mathrm{r}}_2=2\Rightarrow$$$${\mathrm{y}}_{\mathrm{h}}={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{2\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{2\mathrm{x}}\\ {\mathrm{e}}^{\mathrm{x}}{2\mathrm{e}}^{2\mathrm{x}}\end{array}\right|={2\mathrm{e}}^{3\mathrm{x}}-{\mathrm{e}}^{3\mathrm{x}}={\mathrm{e}}^{3\mathrm{x}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}{\mathrm{e}}^{2\mathrm{x}}\\ {4\mathrm{e}}^{2\mathrm{x}}{2\mathrm{e}}^{2\mathrm{x}}\end{array}\right|=-{4\mathrm{e}}^{4\mathrm{x}}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}0\\ {\mathrm{e}}^{\mathrm{x}}{4\mathrm{e}}^{2\mathrm{x}}\end{array}\right|={4\mathrm{e}}^{3\mathrm{x}}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{W}}_1}{\mathrm{W}}\mathrm{dx}=\int \frac{-{4\mathrm{e}}^{4\mathrm{x}}}{{\mathrm{e}}^{3\mathrm{x}}}\mathrm{dx}=-4\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=-{4\mathrm{e}}^{\mathrm{x}}$$$${\mathrm{v}}_2=\int \frac{{\mathrm{W}}_2}{\mathrm{W}}\mathrm{dx}=\int \frac{{4\mathrm{e}}^{3\mathrm{x}}}{{\mathrm{e}}^{3\mathrm{x}}}\mathrm{dx}=4\mathrm{x}\Rightarrow {\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{\mathrm{x}}.(-{4\mathrm{e}}^{\mathrm{x}})+{\mathrm{e}}^{2\mathrm{x}}\left(4\mathrm{x}\right)=-{4\mathrm{e}}^{2\mathrm{x}}+4\mathrm{x}{\mathrm{e}}^{2\mathrm{x}}=(4\mathrm{x}-4){\mathrm{e}}^{2\mathrm{x}}$$$$\mathrm{general}\mathrm{solution}\mathrm{is}\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}$$$$={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{2\mathrm{x}}+(4\mathrm{x}-4){\mathrm{e}}^{2\mathrm{x}}$$$$\mathrm{y}\left(0\right)=-3\Rightarrow \mathrm{a}+\mathrm{b}-4=-3\Rightarrow \mathrm{a}+\mathrm{b}=1$$$${\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)={\mathrm{ae}}^{\mathrm{x}}+2\mathrm{b}{\mathrm{e}}^{2\mathrm{x}}+{4\mathrm{e}}^{2\mathrm{x}}+2(4\mathrm{x}-4){\mathrm{e}}^{2\mathrm{x}}$$$${\mathrm{y}}^{{}^{\prime }}\left(0\right)=5\Rightarrow \mathrm{a}+2\mathrm{b}+4-8=5\Rightarrow \mathrm{a}+2\mathrm{b}=5+4=9\Rightarrow$$$$\mathrm{a}+2(1-\mathrm{a})=9\Rightarrow -\mathrm{a}+2=9\Rightarrow \mathrm{a}=-7$$$$\mathrm{b}=8\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=-{7\mathrm{e}}^{\mathrm{x}}+{8\mathrm{e}}^{2\mathrm{x}}+(4\mathrm{x}-4){\mathrm{e}}^{2\mathrm{x}}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=-{7\mathrm{e}}^{\mathrm{x}}+(4\mathrm{x}+4){\mathrm{e}}^{2\mathrm{x}}$$$$$$

Answered by Ñï= last updated on 08/Mar/21

$$y_p=\frac{1}{D^2-3D+2}4e^{2x}$$$$=4\frac{1}{(D-1)(D-2)}{e}^{2x}$$$$=4e^{2x}(\frac{1}{D}-\frac{1}{D+1})$$$$=4e^{2x}(x-1)$$$$y=C_1{e}^x+C_2{e}^{2x}+4(x-1)e^{2x}$$$$y\left(0\right)=-3y{\left(0\right)}^{\prime }=5$$$$\Rightarrow C_1+2C_2=9$$$$C_1+C_2=1$$$$\Rightarrow C_1=-7C_2=8$$$$y=-7e^x+8e^{2x}+4(x-1)e^{2x}$$$$=-7e^x+(4x+4)e^{2x}$$

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