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Question Number 95839 by mathmax by abdo last updated on 28/May/20

$$\mathrm{solve}{\mathrm{y}}^{\left(3\right)}-{2\mathrm{y}}^{\left(2\right)}+3\mathrm{y}-2\mathrm{y}=\mathrm{sinx}$$

Commented by john santu last updated on 28/May/20

$$\mathrm{homogenous}\mathrm{solution}$$$${\lambda }^3-2{\lambda }^2+3\lambda -2=0$$$$(\lambda -1)({\lambda }^2-\lambda +2)=0$$$$\lambda =1,\frac{1\pm i\sqrt{7}}{2}$$$$y_h=A_1{\mathrm{e}}^x+A_2{\mathrm{e}}^{\frac{1}{2}x}(\cos \left(\frac{\sqrt{7}}{2}\right)x+\sin \left(\frac{\sqrt{7}}{2}\right))$$$$$$

Answered by mathmax by abdo last updated on 29/May/20

$$\mathrm{let}\mathrm{solve}\mathrm{it}\mathrm{by}\mathrm{laplace}\mathrm{transform}$$$$\left(\mathrm{e}\right)\Rightarrow \mathrm{L}({\mathrm{y}}^{\left(3\right))}-2\mathrm{L}\left({\mathrm{y}}^{\left(2\right)}\right)+3\mathrm{L}\left({\mathrm{y}}^{{}^{\prime }}\right)-2\mathrm{L}\left(\mathrm{y}\right)=\mathrm{L}\left(\mathrm{sinx}\right)\Rightarrow$$$${\mathrm{x}}^3\mathrm{L}\left(\mathrm{y}\right)-{\mathrm{x}}^2\mathrm{y}\left(0\right)-{\mathrm{xy}}^{{}^{\prime }}\left(0\right)-{\mathrm{y}}^{{}^{″}}\left(0\right)-2({\mathrm{x}}^2\mathrm{L}\left(\mathrm{y}\right)-\mathrm{xy}\left(0\right)-{\mathrm{y}}^{{}^{\prime }}\left(0\right))$$$$+3(\mathrm{xL}\left(\mathrm{y}\right)-\mathrm{y}\left(0\right))-2\mathrm{L}\left(\mathrm{y}\right)=\mathrm{L}\left(\mathrm{sinx}\right)\Rightarrow$$$$({\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2)\mathrm{L}\left(\mathrm{y}\right)-{\mathrm{x}}^2\mathrm{y}\left(0\right)-{\mathrm{xy}}^{{}^{\prime }}\left(0\right)-{\mathrm{y}}^{\left(2\right)}\left(0\right)+2\mathrm{xy}\left(0\right)+{2\mathrm{y}}^{{}^{\prime }}\left(0\right)$$$$-3\mathrm{y}\left(0\right)=\mathrm{L}\left(\mathrm{sinx}\right)\Rightarrow$$$$({\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2)\mathrm{L}\left(\mathrm{y}\right)=\mathrm{L}\left(\mathrm{sinx}\right)+{\mathrm{x}}^2\mathrm{y}\left(0\right)+(2-\mathrm{x}){\mathrm{y}}^{{}^{\prime }}\left(0\right)+(2\mathrm{x}-3)\mathrm{y}\left(0\right)+{\mathrm{y}}^{\left(2\right)}\left(0\right)\Rightarrow$$$$\mathrm{L}\left(\mathrm{sinx}\right)={\int }_0^{\mathrm{\infty }}\mathrm{sint}{\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{\mathrm{it}-\mathrm{xt}}\mathrm{dt}\right)\mathrm{but}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(\mathrm{i}-\mathrm{x})\mathrm{t}}\mathrm{dt}={\left[\frac{1}{\mathrm{i}-\mathrm{x}}{\mathrm{e}}^{(\mathrm{i}-\mathrm{x})\mathrm{t}}\right]}_0^{\mathrm{\infty }}=-\frac{1}{\mathrm{i}-\mathrm{x}}=\frac{1}{\mathrm{x}-\mathrm{i}}=\frac{\mathrm{x}+\mathrm{i}}{{\mathrm{x}}^2+1}\Rightarrow \mathrm{L}\left(\mathrm{sinx}\right)=\frac{1}{{\mathrm{x}}^2+1}\Rightarrow$$$$\mathrm{L}\left(\mathrm{y}\right)=\frac{1}{({\mathrm{x}}^2+1)({\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2)}+\frac{{\mathrm{x}}^2+2\mathrm{x}-3}{{\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2}\mathrm{y}\left(0\right)+\frac{2-\mathrm{x}}{{\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2}{\mathrm{y}}^{{}^{\prime }}\left(0\right)+\frac{{\mathrm{y}}^{\left(2\right)}\left(0\right)}{{\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2}$$$$\Rightarrow \mathrm{y}={\mathrm{L}}^{-1}\left(\frac{1}{({\mathrm{x}}^2+1)({\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2)}\right)+{\mathrm{y}}_0{\mathrm{L}}^{-1}\left(\frac{{\mathrm{x}}^2+2\mathrm{x}-3}{{\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2)}\right)$$$$+{\mathrm{y}}^{{}^{\prime }}\left(0\right){\mathrm{L}}^{-1}\left(\frac{2-\mathrm{x}}{{\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2}\right)+{\mathrm{y}}^{\left(2\right)}\left(0\right){\mathrm{L}}^{-1}\left(\frac{1}{{\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2}\right)\mathrm{wehave}$$$${\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2={\mathrm{x}}^3-{\mathrm{x}}^2-{\mathrm{x}}^2+3\mathrm{x}-2={\mathrm{x}}^2(\mathrm{x}-1)-({\mathrm{x}}^2-3\mathrm{x}+2)$$$$={\mathrm{x}}^2(\mathrm{x}-1)-({\mathrm{x}}^2-\mathrm{x}-2\mathrm{x}+2)={\mathrm{x}}^2(\mathrm{x}-1)-\{\mathrm{x}(\mathrm{x}-1)-2(\mathrm{x}-1)\}$$$$={\mathrm{x}}^2(\mathrm{x}-1)-(\mathrm{x}-1)(\mathrm{x}-2)=(\mathrm{x}-1)({\mathrm{x}}^2-\mathrm{x}+2)\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^3-{2\mathrm{x}}^2+3\mathrm{x}-2}=\frac{1}{(\mathrm{x}-1)({\mathrm{x}}^2-\mathrm{x}+2)}=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{\mathrm{bx}+\mathrm{c}}{{\mathrm{x}}^2-\mathrm{x}+2}$$$$....\mathrm{be}\mathrm{continued}....$$

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