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Question Number 96773 by abdomathmax last updated on 04/Jun/20

$$\mathrm{solve}{\mathrm{y}}^{{}^{″}}-\mathrm{y}=\frac{\mathrm{sinx}}{\mathrm{x}}$$

Answered by abdomathmax last updated on 05/Jun/20

$$\mathrm{let}\mathrm{solve}\mathrm{by}\mathrm{laplace}$$$$\left(\mathrm{e}\right)\Rightarrow \mathrm{L}\left({\mathrm{y}}^{{}^{″}}\right)-\mathrm{L}\left(\mathrm{y}\right)=\mathrm{L}\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)\Rightarrow$$$${\mathrm{x}}^2\mathrm{L}\left(\mathrm{y}\right)-\mathrm{xy}\left(0\right)-{\mathrm{y}}^{{}^{\prime }}\left(0\right)-\mathrm{L}\left(\mathrm{y}\right)=\mathrm{L}\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)\Rightarrow$$$$({\mathrm{x}}^2-1)\mathrm{L}\left(\mathrm{y}\right)=\mathrm{xy}\left(0\right)+{\mathrm{y}}^{{}^{\prime }}\left(0\right)+\mathrm{L}\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)$$$$\mathrm{we}\mathrm{have}\mathrm{L}\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sint}}{\mathrm{t}}{\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}=\mathrm{f}\left(\mathrm{x}\right)$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=-{\int }_0^{\mathrm{\infty }}\mathrm{sint}{\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}=-\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{\mathrm{it}-\mathrm{xt}}\mathrm{dt}\right)$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(-\mathrm{x}+\mathrm{i})\mathrm{t}}\mathrm{dt}={\left[\frac{1}{-\mathrm{x}+\mathrm{i}}{\mathrm{e}}^{(-\mathrm{x}+\mathrm{i})\mathrm{t}}\right]}_0^{\mathrm{\infty }}$$$$=-\frac{1}{-\mathrm{x}+\mathrm{i}}=\frac{1}{\mathrm{x}-\mathrm{i}}=\frac{\mathrm{x}+\mathrm{i}}{{\mathrm{x}}^2+1}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=-\frac{1}{1+{\mathrm{x}}^2}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=-\mathrm{arrctanx}+\mathrm{c}$$$$\mathrm{f}\left(0\right)=\frac{\pi }{2}=0+\mathrm{c}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\pi }{2}-\mathrm{arcrtanx}$$$$\mathrm{e}\Rightarrow ({\mathrm{x}}^2-1)\mathrm{L}\left(\mathrm{y}\right)=\mathrm{xy}\left(0\right)+{\mathrm{y}}^{{}^{\prime }}\left(0\right)+\frac{\pi }{2}-\mathrm{arctanx}$$$$\Rightarrow \mathrm{L}\left(\mathrm{y}\right)=\mathrm{y}\left(0\right)\frac{\mathrm{x}}{{\mathrm{x}}^2-1}+\frac{{\mathrm{y}}^{{}^{\prime }}\left(0\right)+\frac{\pi }{2}}{{\mathrm{x}}^2-1}-\frac{\mathrm{arctanx}}{{\mathrm{x}}^2-1}$$$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{y}\left(0\right){\mathrm{L}}^{-1}\left(\frac{\mathrm{x}}{{\mathrm{x}}^2-1}\right)+({\mathrm{y}}^{{}^{\prime }}\left(0\right)+\frac{\pi }{2}){\mathrm{L}}^{-1}\left(\frac{1}{{\mathrm{x}}^2-1}\right)$$$$-{\mathrm{L}}^{-1}\left(\frac{\mathrm{arctanx}}{{\mathrm{x}}^2-1}\right)$$$$\frac{\mathrm{x}}{{\mathrm{x}}^2-1}=\frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}+1)}=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{\mathrm{b}}{\mathrm{x}+1}\Rightarrow$$$$\mathrm{a}=\frac{1}{2}\mathrm{and}\mathrm{b}=\frac{1}{2}\Rightarrow {\mathrm{L}}^{-1}\left(\frac{\mathrm{x}}{{\mathrm{x}}^2-1}\right)$$$$=\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}+\frac{1}{2}{\mathrm{e}}^{-\mathrm{x}}=\mathrm{ch}\left(\mathrm{x}\right)$$$$\frac{1}{{\mathrm{x}}^2-1}=\frac{1}{2}(\frac{1}{\mathrm{x}-1}-\frac{1}{\mathrm{x}+1})\Rightarrow$$$${\mathrm{L}}^{-1}\left(\frac{1}{{\mathrm{x}}^2-1}\right)=\frac{{\mathrm{e}}^{\mathrm{x}}}{2}-\frac{{\mathrm{e}}^{-\mathrm{x}}}{2}=\mathrm{sh}\left(\mathrm{x}\right)\mathrm{rest}\mathrm{to}\mathrm{find}$$$${\mathrm{L}}^{-1}\left(\frac{\mathrm{arctanx}}{{\mathrm{x}}^2-1}\right)=\delta \left(\mathrm{x}\right)\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{y}\left(\mathrm{o}\right)\mathrm{ch}\left(\mathrm{x}\right)+({\mathrm{y}}^{{}^{\prime }}\left(0\right)+\frac{\pi }{2})\mathrm{shx}+\delta \left(\mathrm{x}\right)$$$$$$

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