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Question Number 127152 by mnjuly1970 last updated on 27/Dec/20

$$...differentialequation...$$ $$if\frac{d^{2}y}{{dx}^2}=y\left(x\right)\mathrm{\&}y\left(0\right)=-y^{\prime }\left(0\right)=1$$ $$thenevaluate:::$$ $$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}y\left(x\right).y^{-1}(-x)dx=?$$ $$$$

Answered by mathmax by abdo last updated on 27/Dec/20

$${\mathrm{y}}^{{}^{″}}=\mathrm{y}\mathrm{and}\mathrm{y}\left(0\right)=1,{\mathrm{y}}^{{}^{\prime }}\left(0\right)=-1$$ $${\mathrm{r}}^2-1=0\Rightarrow \mathrm{y}={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{-\mathrm{x}}$$ $$\mathrm{y}\left(0\right)=1\Rightarrow \mathrm{a}+\mathrm{b}=1,{\mathrm{y}}^{{}^{\prime }}={\mathrm{ae}}^{\mathrm{x}}-{\mathrm{be}}^{-\mathrm{x}},{\mathrm{y}}^{{}^{\prime }}\left(0\right)=-1\Rightarrow \mathrm{a}-\mathrm{b}=-1\Rightarrow$$ $$\{\begin{array}{l}\mathrm{a}+\mathrm{b}=1\\ \mathrm{a}-\mathrm{b}=-1\Rightarrow \{\begin{array}{l}\mathrm{a}=0\Rightarrow \mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{b}=1\end{array}\end{array}$$ $$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{t}\Rightarrow \mathrm{x}={\mathrm{y}}^{-1}\left(\mathrm{t}\right)\Rightarrow {\mathrm{e}}^{-\mathrm{x}}=\mathrm{t}\Rightarrow -\mathrm{x}=\ln \left(\mathrm{t}\right)\Rightarrow \mathrm{x}=-\mathrm{lnt}\Rightarrow$$ $${\mathrm{y}}^{-1}\left(\mathrm{x}\right)=-\mathrm{lnx}\Rightarrow \mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}(-\ln (-\mathrm{x}))\mathrm{dx}$$ $$=-{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}(\mathrm{i}\pi +\ln \left(\mathrm{x}\right))\mathrm{dx}=-\mathrm{i}\pi {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}-{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\ln \left(\mathrm{x}\right)\mathrm{dx}$$ $$=-\mathrm{i}\pi {[-{\mathrm{e}}^{-\mathrm{x}}]}_0^{\mathrm{\infty }}+\gamma =\gamma -\mathrm{i}\pi$$

Commented bymnjuly1970 last updated on 27/Dec/20

$$thanksalotsirmax...$$

Commented bymathmax by abdo last updated on 27/Dec/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Answered by mnjuly1970 last updated on 27/Dec/20

$$y^{″}-y=0\Rightarrow m^2-1=0$$ $$m=\pm 1\Rightarrow y\left(x\right)=c_1{e}^x+c_2{e}^{-x}$$ $$usingtheinitialconditions::$$ $$\{\begin{array}{l}{c}_1+c_2=1\\ c_1-c_2=-1\end{array}\{\begin{array}{l}{c}_1=0\\ c_2=1\end{array}$$ $$y\left(x\right)=e^{-x}$$ $$y(-x)=e^x\Rightarrow y^{-1}(-x)=ln\left(x\right)$$ $${\int }_0^{\mathrm{\infty }}{e}^{-x}ln\left(x\right)=-\gamma ...✓$$

Answered by Dwaipayan Shikari last updated on 27/Dec/20

$$\frac{d^{2}y}{{dx}^2}=y\left(x\right)\Rightarrow y\left(x\right)=\mathrm{\Lambda }{e}^x+\mathrm{\Psi }{e}^{-x}$$ $$y\left(0\right)=\mathrm{\Lambda }+\mathrm{\Psi }=1$$ $$y^{\prime }\left(0\right)=\mathrm{\Lambda }-\mathrm{\Psi }=-1\mathrm{\Lambda }=0,\mathrm{\Psi }=1$$ $$y\left(x\right)=e^{-x}\Rightarrow -x=log\left(y\right)$$ $$y^{-1}(-x)=log\left(x\right)$$ $${\int }_0^{\mathrm{\infty }}{e}^{-x}log\left(x\right)dx=-\gamma$$

Commented bymnjuly1970 last updated on 27/Dec/20

$$grateful...$$

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