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

...differential equation... if (d^2 y/dx^2 ) = y(x) & y(0)=−y′(0)=1 then evaluate ::: Ω=∫_0 ^( ∞) y(x).y^(−1) (−x)dx=?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{differential}\:\:{equation}...\:\:\:\:\:\:\:\: \\ $$ $$\:\:\:\:\:{if}\:\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:{y}\left({x}\right)\:\:\&\:{y}\left(\mathrm{0}\right)=−{y}'\left(\mathrm{0}\right)=\mathrm{1} \\ $$ $$\:\:{then}\:\:\:{evaluate}\:\:::: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} {y}\left({x}\right).{y}^{−\mathrm{1}} \left(−{x}\right){dx}=? \\ $$ $$ \\ $$

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

y^(′′) =y and y(0)=1 ,y^′ (0)=−1 r^2 −1=0 ⇒y =ae^x +be^(−x) y(0)=1 ⇒a+b=1 ,y^′ =ae^x −be^(−x) ,y^′ (0)=−1 ⇒a−b=−1 ⇒ { ((a+b=1)),((a−b=−1 ⇒ { ((a=0 ⇒y(x)=e^(−x) )),((b=1 )) :})) :} y(x)=t ⇒x=y^(−1) (t) ⇒e^(−x) =t ⇒−x =ln(t) ⇒x=−lnt ⇒ y^(−1) (x)=−lnx ⇒Ω =∫_0 ^∞ e^(−x) (−ln(−x))dx =−∫_0 ^∞ e^(−x) (iπ +ln(x))dx =−iπ ∫_0 ^∞ e^(−x) dx −∫_0 ^∞ e^(−x) ln(x)dx =−iπ[−e^(−x) ]_0 ^∞ +γ =γ−iπ

$$\mathrm{y}^{''} =\mathrm{y}\:\:\mathrm{and}\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1}\:,\mathrm{y}^{'} \left(\mathrm{0}\right)=−\mathrm{1} \\ $$ $$\mathrm{r}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{y}\:=\mathrm{ae}^{\mathrm{x}} \:+\mathrm{be}^{−\mathrm{x}} \\ $$ $$\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1}\:\Rightarrow\mathrm{a}+\mathrm{b}=\mathrm{1}\:,\mathrm{y}^{'} \:=\mathrm{ae}^{\mathrm{x}} −\mathrm{be}^{−\mathrm{x}} \:,\mathrm{y}^{'} \left(\mathrm{0}\right)=−\mathrm{1}\:\Rightarrow\mathrm{a}−\mathrm{b}=−\mathrm{1}\:\Rightarrow \\ $$ $$\begin{cases}{\mathrm{a}+\mathrm{b}=\mathrm{1}}\\{\mathrm{a}−\mathrm{b}=−\mathrm{1}\:\:\Rightarrow\begin{cases}{\mathrm{a}=\mathrm{0}\:\Rightarrow\mathrm{y}\left(\mathrm{x}\right)=\mathrm{e}^{−\mathrm{x}} }\\{\mathrm{b}=\mathrm{1}\:\:}\end{cases}}\end{cases} \\ $$ $$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{t}\:\Rightarrow\mathrm{x}=\mathrm{y}^{−\mathrm{1}} \left(\mathrm{t}\right)\:\Rightarrow\mathrm{e}^{−\mathrm{x}} \:=\mathrm{t}\:\Rightarrow−\mathrm{x}\:=\mathrm{ln}\left(\mathrm{t}\right)\:\Rightarrow\mathrm{x}=−\mathrm{lnt}\:\Rightarrow \\ $$ $$\mathrm{y}^{−\mathrm{1}} \left(\mathrm{x}\right)=−\mathrm{lnx}\:\Rightarrow\Omega\:=\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{x}} \:\left(−\mathrm{ln}\left(−\mathrm{x}\right)\right)\mathrm{dx} \\ $$ $$=−\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}} \left(\mathrm{i}\pi\:+\mathrm{ln}\left(\mathrm{x}\right)\right)\mathrm{dx}\:\:=−\mathrm{i}\pi\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}} \mathrm{dx}\:−\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{dx} \\ $$ $$=−\mathrm{i}\pi\left[−\mathrm{e}^{−\mathrm{x}} \right]_{\mathrm{0}} ^{\infty} +\gamma\:\:=\gamma−\mathrm{i}\pi \\ $$

Commented bymnjuly1970 last updated on 27/Dec/20

thanks alot sir max...

$${thanks}\:{alot}\:{sir}\:\:{max}... \\ $$

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

you are welcome

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

Answered by mnjuly1970 last updated on 27/Dec/20

y′′−y=0 ⇒ m^2 −1=0 m=±1⇒ y(x)=c_1 e^x +c_2 e^(−x) using the initial conditions:: { ((c_1 +c_2 =1)),((c_1 −c_2 =−1)) :} { ((c_1 =0)),((c_2 =1)) :} y(x)=e^(−x) y(−x)=e^x ⇒y^(−1) (−x)=ln(x) ∫_(0 ) ^( ∞) e^(−x) ln(x) = −γ ...✓

$${y}''−{y}=\mathrm{0}\:\Rightarrow\:{m}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$ $$\:{m}=\pm\mathrm{1}\Rightarrow\:{y}\left({x}\right)={c}_{\mathrm{1}} {e}^{{x}} +{c}_{\mathrm{2}} {e}^{−{x}} \\ $$ $$\:{using}\:{the}\:{initial}\:{conditions}:: \\ $$ $$\:\:\begin{cases}{{c}_{\mathrm{1}} +{c}_{\mathrm{2}} =\mathrm{1}}\\{{c}_{\mathrm{1}} −{c}_{\mathrm{2}} =−\mathrm{1}}\end{cases}\begin{cases}{{c}_{\mathrm{1}} =\mathrm{0}}\\{{c}_{\mathrm{2}} =\mathrm{1}}\end{cases} \\ $$ $$\:\:{y}\left({x}\right)={e}^{−{x}} \\ $$ $${y}\left(−{x}\right)={e}^{{x}} \:\:\Rightarrow{y}^{−\mathrm{1}} \left(−{x}\right)={ln}\left({x}\right) \\ $$ $$\:\int_{\mathrm{0}\:} ^{\:\infty} {e}^{−{x}} {ln}\left({x}\right)\:=\:−\gamma\:\:\:\:...\checkmark \\ $$

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

(d^2 y/dx^2 )=y(x) ⇒y(x)=Λe^x +Ψe^(−x) y(0)=Λ+Ψ=1 y′(0)=Λ−Ψ=−1 Λ=0 ,Ψ=1 y(x)=e^(−x) ⇒−x=log(y) y^(−1) (−x)=log(x) ∫_0 ^∞ e^(−x) log(x)dx =−γ

$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }={y}\left({x}\right)\:\:\Rightarrow{y}\left({x}\right)=\Lambda{e}^{{x}} +\Psi{e}^{−{x}} \\ $$ $${y}\left(\mathrm{0}\right)=\Lambda+\Psi=\mathrm{1} \\ $$ $${y}'\left(\mathrm{0}\right)=\Lambda−\Psi=−\mathrm{1}\:\:\:\:\:\:\:\Lambda=\mathrm{0}\:\:,\Psi=\mathrm{1} \\ $$ $${y}\left({x}\right)={e}^{−{x}} \Rightarrow−{x}={log}\left({y}\right) \\ $$ $${y}^{−\mathrm{1}} \left(−{x}\right)={log}\left({x}\right) \\ $$ $$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {log}\left({x}\right){dx}\:\:\:\:\:\:=−\gamma \\ $$

Commented bymnjuly1970 last updated on 27/Dec/20

grateful...

$${grateful}... \\ $$

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