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Question Number 106319 by mohammad17 last updated on 04/Aug/20

Commented by mohammad17 last updated on 04/Aug/20

$$byusinglaplasetransformesolvethisequation?$$

Commented by rexfordattacudjoe last updated on 04/Aug/20

$$x^3+2x^2-3x=0$$$$x^2+2x-3=0$$$$x=1orx=-3orx=0$$$$y=c_1{e}^x+c_2{e}^{-3x}+c_3$$$$y\left(0\right)=4,c_1+c_2+c_3=4........\left(1\right)$$$$y^{\prime }\left(0\right)=8,c_1{e}^x-3c_2{e}^{-3x}=y^{\prime }$$$$c_1-3c_2=8..........\left(2\right)$$$$y^{″}\left(0\right)=-4,c_1{e}^x+9c_2{e}^{-3x}=y^{″}$$$$c_1+9c_2=-4.......\left(3\right)$$$$c_1=5,c_2=-1and{c}_3=0$$$$\therefore y=5e^x-e^{-3x}$$

Answered by mathmax by abdo last updated on 04/Aug/20

$${\mathrm{y}}^{\left(3\right)}+{2\mathrm{y}}^{\left(2\right)}-{3\mathrm{y}}^{\left(1\right)}=0\Rightarrow$$$$\mathrm{L}\left({\mathrm{y}}^{\left(3\right)}\right)+2\mathrm{L}\left({\mathrm{y}}^{\left(2\right)}\right)-3\mathrm{L}\left({\mathrm{y}}^{\left(1\right)}\right)=0\Rightarrow$$$${\mathrm{x}}^3\mathrm{L}\left(\mathrm{y}\right)-{\mathrm{x}}^2\mathrm{y}\left(0\right)-{\mathrm{xy}}^{{}^{\prime }}\left(0\right)-{\mathrm{y}}^{{}^{″}}\left(0\right)+2({\mathrm{x}}^2\mathrm{L}\left(\mathrm{y}\right)-\mathrm{xy}\left(\mathrm{o}\right)-{\mathrm{y}}^{{}^{\prime }}\left(0\right))$$$$-3(\mathrm{xL}\left(\mathrm{y}\right)-\mathrm{y}\left(0\right)=0\Rightarrow$$$${\mathrm{x}}^3\mathrm{L}\left(\mathrm{y}\right)-{4\mathrm{x}}^2-8\mathrm{x}+4+2({\mathrm{x}}^2\mathrm{L}\left(\mathrm{y}\right)-4\mathrm{x}-8)-3(\mathrm{xL}\left(\mathrm{y}\right)-4)=0\Rightarrow$$$$({\mathrm{x}}^3+{2\mathrm{x}}^2-3\mathrm{x})\mathrm{L}\left(\mathrm{y}\right)-{4\mathrm{x}}^2-8\mathrm{x}+4-8\mathrm{x}-16+12=0\Rightarrow$$$$({\mathrm{x}}^3+{2\mathrm{x}}^2-3\mathrm{x})\mathrm{L}\left(\mathrm{y}\right)-{4\mathrm{x}}^2-16\mathrm{x}=0\Rightarrow$$$$({\mathrm{x}}^2+2\mathrm{x}-3)\mathrm{L}\left(\mathrm{y}\right)-4\mathrm{x}-16=0\Rightarrow \mathrm{L}\left(\mathrm{y}\right)=\frac{4\mathrm{x}+16}{{\mathrm{x}}^2+2\mathrm{x}-3}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{L}}^{-1}\left(\frac{4\mathrm{x}+16}{{\mathrm{x}}^2+2\mathrm{x}-3}\right)\mathrm{let}\mathrm{decompose}\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}+16}{{\mathrm{x}}^2+2\mathrm{x}-3}$$$${\mathrm{x}}^2+2\mathrm{x}-3=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1+3=4\Rightarrow {\mathrm{x}}_1=-1+2=1$$$${\mathrm{x}}_2=-1-2=-3\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}+16}{(\mathrm{x}-1)(\mathrm{x}+3)}=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{\mathrm{b}}{\mathrm{x}+3}$$$$\mathrm{a}=\frac{20}{4}=5\mathrm{and}\mathrm{b}=\frac{4}{-4}=-1\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{5}{\mathrm{x}-1}-\frac{1}{\mathrm{x}+3}\Rightarrow$$$${\mathrm{L}}^{-1}\left(\mathrm{f}\right)={5\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-3\mathrm{x}}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)={5\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-3\mathrm{x}}$$

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