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Question Number 98993 by Rio Michael last updated on 17/Jun/20

$$\mathrm{Use}\mathrm{the}\mathrm{laplace}\mathrm{tranform}\mathrm{to}\mathrm{solve}\frac{d^{2}y}{{dx}^2}+5\frac{dy}{dx}+6y=e^{-x}$$$$\mathrm{for}y=0,\mathrm{and}\frac{dy}{dx}=1\mathrm{when}x=0$$

Answered by mathmax by abdo last updated on 18/Jun/20

$${\mathrm{y}}^{{}^{″}}+{5\mathrm{y}}^{{}^{\prime }}+6\mathrm{y}={\mathrm{e}}^{-\mathrm{x}}\mathrm{with}\mathrm{y}\left(0\right)=0\mathrm{and}{\mathrm{y}}^{{}^{\prime }}\left(0\right)=1$$$$\left(\mathrm{e}\right)\Rightarrow \mathrm{L}\left({\mathrm{y}}^{{}^{″}}\right)+5\mathrm{L}\left({\mathrm{y}}^{{}^{\prime }}\right)+6\mathrm{L}\left(\mathrm{y}\right)=\mathrm{L}\left({\mathrm{e}}^{-\mathrm{x}}\right)\Rightarrow$$$${\mathrm{x}}^2\mathrm{L}\left(\mathrm{y}\right)-\mathrm{x}\mathrm{y}\left(0\right)-{\mathrm{y}}^{{}^{\prime }}\left(0\right)+5(\mathrm{x}\mathrm{L}\left(\mathrm{y}\right)-\mathrm{y}\left(0\right))+6\mathrm{L}\left(\mathrm{y}\right)=\mathrm{L}\left({\mathrm{e}}^{-\mathrm{x}}\right)\Rightarrow$$$$({\mathrm{x}}^2+5\mathrm{x}+6)\mathrm{L}\left(\mathrm{y}\right)-1=\mathrm{L}\left({\mathrm{e}}^{-\mathrm{x}}\right)\mathrm{we}\mathrm{have}\mathrm{L}\left({\mathrm{e}}^{-\mathrm{x}}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}{\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{x}+1)\mathrm{t}}\mathrm{dt}={[-\frac{1}{\mathrm{x}+1}{\mathrm{e}}^{-(\mathrm{x}+1)\mathrm{t}}]}_0^{\mathrm{\infty }}=\frac{1}{\mathrm{x}+1}$$$$\left(\mathrm{e}\right)\Rightarrow ({\mathrm{x}}^2+5\mathrm{x}+6)\mathrm{L}\left(\mathrm{y}\right)=1+\frac{1}{\mathrm{x}+1}=\frac{\mathrm{x}+2}{\mathrm{x}+1}\Rightarrow \mathrm{L}\left(\mathrm{y}\right)=\frac{\mathrm{x}+2}{(\mathrm{x}+1)({\mathrm{x}}^2+5\mathrm{x}+6)}\Rightarrow$$$$\mathrm{y}={\mathrm{L}}^{-1}\left(\frac{\mathrm{x}+2}{(\mathrm{x}+1)({\mathrm{x}}^2+5\mathrm{x}+6)}\right)\mathrm{let}\mathrm{decompose}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+2}{(\mathrm{x}+1)({\mathrm{x}}^2+5\mathrm{x}+6)}$$$${\mathrm{x}}^2+5\mathrm{x}+6=0\to \mathrm{\Delta }=25-24=1\Rightarrow {\mathrm{x}}_1=\frac{-5+1}{2}=-2\mathrm{and}{\mathrm{x}}_2=\frac{-5-1}{2}=-3\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+2}{(\mathrm{x}+1)(\mathrm{x}+2)(\mathrm{x}+3)}=\frac{1}{(\mathrm{x}+1)(\mathrm{x}+3)}=\frac{1}{2}(\frac{1}{\mathrm{x}+1}-\frac{1}{\mathrm{x}+3})\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{L}}^{-1}\left(\mathrm{f}\right)=\frac{1}{2}{\mathrm{L}}^{-1}\left(\frac{1}{\mathrm{x}+1}\right)-\frac{1}{2}{\mathrm{L}}^{-1}\left(\frac{1}{\mathrm{x}+3}\right)=\frac{1}{2}{\mathrm{e}}^{-\mathrm{x}}-\frac{1}{2}{\mathrm{e}}^{-3\mathrm{x}}$$$$\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{y}\left(\mathrm{x}\right)=\frac{{\mathrm{e}}^{-\mathrm{x}}-{\mathrm{e}}^{-3\mathrm{x}}}{2}$$

Commented by Rio Michael last updated on 18/Jun/20

$$\mathrm{wow}\mathrm{nice}\mathrm{solution}\mathrm{sir},\mathrm{but}\mathrm{i}\mathrm{have}\mathrm{problems}\mathrm{with}$$$$\mathrm{L}\left(y^{″}\right)\mathrm{and}\mathrm{L}\left(y^{″}\right)\mathrm{the}\mathrm{derivations}$$

Commented by abdomathmax last updated on 18/Jun/20

$$\mathrm{you}\mathrm{must}\mathrm{use}\mathrm{a}\mathrm{table}\mathrm{of}\mathrm{formula}\mathrm{for}\mathrm{laplace}$$

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