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Question Number 105266 by mohammad17 last updated on 27/Jul/20

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

$$\mathrm{we}\mathrm{have}\mathrm{L}\left({\mathrm{t}}^{\frac{3}{2}}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\frac{3}{2}}{\mathrm{e}}^{-\mathrm{tx}}\mathrm{dx}{=}_{\mathrm{tx}=\mathrm{u}}{\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{u}}{\mathrm{t}}\right)}^{\frac{3}{2}}{\mathrm{e}}^{-\mathrm{u}}\frac{\mathrm{du}}{\mathrm{t}}$$$$=\frac{1}{{\mathrm{t}}^{\frac{3}{2}+1}}{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^{\frac{3}{2}}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}=\frac{1}{{\mathrm{t}}^{\frac{5}{2}}}{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^{\frac{5}{2}-1}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}=\frac{\mathrm{\Gamma }\left(\frac{5}{2}\right)}{{\mathrm{t}}^{\frac{5}{2}}}$$$$\mathrm{\Gamma }\left(\frac{5}{2}\right)=\mathrm{\Gamma }(\frac{3}{2}+1)=\frac{3}{2}\mathrm{\Gamma }\left(\frac{3}{2}\right)=\frac{3}{2}\mathrm{\Gamma }(\frac{1}{2}+1)=\frac{3}{4}\mathrm{\Gamma }\left(\frac{1}{2}\right)$$$$\mathrm{\Gamma }\left(\frac{1}{2}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{1}{2}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}}{\sqrt{\mathrm{t}}}\mathrm{dt}{=}_{\sqrt{\mathrm{t}}=\mathrm{u}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{u}}^2}}{\mathrm{u}}\left(2\mathrm{u}\right)\mathrm{du}$$$$=2{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}=\sqrt{\pi }\Rightarrow \mathrm{L}\left({\mathrm{t}}^{\frac{3}{2}}\right)=\frac{3}{4}\times \frac{\sqrt{\pi }}{{\mathrm{t}}^{\frac{5}{2}}}$$

Answered by abdomathmax last updated on 27/Jul/20

$$\mathrm{L}\left(\mathrm{h}\left(\mathrm{s}\right)\right)={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^2\sin \left(2\mathrm{t}\right){\mathrm{e}}^{-\mathrm{st}}\mathrm{dt}$$$$=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{t}}^2{\mathrm{e}}^{2\mathrm{it}-\mathrm{st}}\mathrm{dt}\right)=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{t}}^2{\mathrm{e}}^{(2\mathrm{i}-\mathrm{s})\mathrm{t}}\mathrm{dt}\right)\mathrm{but}\mathrm{by}\mathrm{parts}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{t}}^2{\mathrm{e}}^{(2\mathrm{i}-\mathrm{s})\mathrm{t}}\mathrm{dt}={\left[\frac{{\mathrm{t}}^2}{2\mathrm{i}-\mathrm{s}}{\mathrm{e}}^{(2\mathrm{i}-\mathrm{s})\mathrm{t}}\right]}_{\mathrm{t}=0}^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}2\mathrm{t}.\frac{1}{2\mathrm{i}-\mathrm{s}}{\mathrm{e}}^{(2\mathrm{i}-\mathrm{s})\mathrm{t}}\mathrm{dt}$$$$=-\frac{2}{2\mathrm{i}-\mathrm{s}}{\int }_0^{\mathrm{\infty }}\mathrm{t}{\mathrm{e}}^{(2\mathrm{i}-\mathrm{s})\mathrm{t}}\mathrm{dt}$$$$=\frac{-2}{2\mathrm{i}-\mathrm{s}}\{{\left[\frac{\mathrm{t}}{2\mathrm{i}-\mathrm{s}}{\mathrm{e}}^{(2\mathrm{i}-\mathrm{s})\mathrm{t}}\right]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\frac{1}{2\mathrm{i}-\mathrm{s}}{\mathrm{e}}^{(2\mathrm{i}-\mathrm{s})\mathrm{t}}\mathrm{dt}\}$$$$=\frac{2}{{(2\mathrm{i}-\mathrm{s})}^2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(2\mathrm{i}-\mathrm{s})\mathrm{t}}\mathrm{dt}=\frac{2}{{(2\mathrm{i}-\mathrm{s})}^2}{\left[\frac{1}{2\mathrm{i}-\mathrm{s}}{\mathrm{e}}^{(2\mathrm{i}-\mathrm{s})\mathrm{t}}\right]}_0^{\mathrm{\infty }}$$$$=-\frac{2}{{(2\mathrm{i}-\mathrm{s})}^3}=\frac{2}{{(\mathrm{s}-2\mathrm{i})}^3}=\frac{2{(\mathrm{s}+2\mathrm{i})}^3}{{({\mathrm{s}}^2+4)}^3}$$$$=\frac{2({\mathrm{s}}^3+{3\mathrm{s}}^2\left(2\mathrm{i}\right)+3\mathrm{s}{\left(2\mathrm{i}\right)}^2+{\left(2\mathrm{i}\right)}^3)}{{({\mathrm{s}}^2+4)}^3}$$$$=\frac{2({\mathrm{s}}^3+{6\mathrm{is}}^2-12\mathrm{s}-8\mathrm{i})}{{({\mathrm{s}}^2+4)}^3}\Rightarrow$$$$\mathrm{Im}(...)=\frac{2({6\mathrm{s}}^2-8)}{{({\mathrm{s}}^2+4)}^3}\Rightarrow \mathrm{F}\left(\mathrm{s}\right)=\frac{4({3\mathrm{s}}^2-4)}{{({\mathrm{s}}^2+4)}^3}$$$$$$

Answered by Aziztisffola last updated on 27/Jul/20

$$\mathrm{A})\mathrm{F}\left(\mathrm{s}\right)={(-1)}^2\frac{{\mathrm{d}}^2}{{\mathrm{ds}}^2}\mathrm{L}\left(\sin 2\mathrm{t}\right)=\frac{{\mathrm{d}}^2}{{\mathrm{ds}}^2}\mathrm{L}\left(\sin 2\mathrm{t}\right)$$$$=\frac{{\mathrm{d}}^2}{{\mathrm{ds}}^2}\left(\frac{2}{{\mathrm{s}}^2+4}\right)=\frac{\mathrm{d}}{\mathrm{ds}}\left(\frac{-4\mathrm{s}}{{({\mathrm{s}}^2+4)}^2}\right)=\frac{-4{({\mathrm{s}}^2+4)}^2+4\mathrm{s}.2\mathrm{s}({\mathrm{s}}^2+4)}{{({\mathrm{s}}^2+4)}^4}$$$$=\frac{-4({\mathrm{s}}^2+4)+{8\mathrm{s}}^2}{{({\mathrm{s}}^2+4)}^3}=\frac{{4\mathrm{s}}^2-16}{{({\mathrm{s}}^2+4)}^3}$$$$\mathrm{F}\left(\mathrm{s}\right)=\frac{{4\mathrm{s}}^2-16}{{({\mathrm{s}}^2+4)}^3}$$

Answered by bemath last updated on 27/Jul/20

Answered by Aziztisffola last updated on 27/Jul/20

$$\left(\mathrm{B}\right)\mathrm{L}\left({\mathrm{t}}^{\frac{3}{2}}\right)=\mathrm{L}\left(\mathrm{t}\sqrt{\mathrm{t}}\right)=-\frac{\mathrm{d}}{\mathrm{ds}}\mathrm{L}\left(\sqrt{\mathrm{t}}\right)=-{\left(\frac{\sqrt{\pi }}{2\sqrt{{\mathrm{s}}^3}}\right)}^{\prime }$$$$=-\frac{\sqrt{\pi }}{2}{\left(\frac{1}{\sqrt{{\mathrm{s}}^3}}\right)}^{\prime }=\frac{3\sqrt{\pi }}{4}\frac{\mathrm{s}}{\sqrt{\mathrm{s}}}=\frac{3\sqrt{\pi }}{4}\sqrt{\mathrm{s}}$$$$\mathrm{L}\left({\mathrm{t}}^{\frac{3}{2}}\right)=\frac{3\sqrt{\pi }}{4}\sqrt{\mathrm{s}}$$

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

$$\mathrm{b})\mathrm{x}(\mathrm{y}+2){\mathrm{y}}^{{}^{\prime }}=\mathrm{lnx}+1\mathrm{andy}\left(1\right)=-1$$$$(\mathrm{y}+2){\mathrm{y}}^{{}^{\prime }}=\frac{\mathrm{lnx}+1}{\mathrm{x}}\Rightarrow \int (\mathrm{y}+2){\mathrm{y}}^{{}^{\prime }}\mathrm{dy}=\int \frac{\mathrm{lnx}+1}{\mathrm{x}}\mathrm{dx}\Rightarrow$$$$\frac{1}{2}{\mathrm{y}}^2+2\mathrm{y}=\ln \mid \mathrm{x}\mid +\int \frac{\mathrm{lnx}}{\mathrm{x}}\mathrm{dx}\Rightarrow {\mathrm{y}}^2+4\mathrm{y}-2\{\ln \mid \mathrm{x}\mid +\int \frac{\mathrm{lnx}}{\mathrm{x}}\mathrm{dx}\}=0$$$${\mathrm{\Delta }}^{{}^{\prime }}=4+2\{\ln \mid \mathrm{x}\mid +\int \frac{\mathrm{lnx}}{\mathrm{x}}\mathrm{dx}\}\Rightarrow \mathrm{y}=-2\stackrel{-}{+}\sqrt{\ln \mid \mathrm{x}\mid +\int \frac{\mathrm{lnx}}{\mathrm{x}}\mathrm{dx}}$$

Commented by abdomathmax last updated on 27/Jul/20

$$\mathrm{but}\int \frac{\mathrm{lnx}}{\mathrm{x}}\mathrm{dx}=\frac{1}{2}{\ln }^2\mathrm{x}+\mathrm{c}\Rightarrow$$$$\mathrm{y}=-2\stackrel{-}{+}\sqrt{\ln \mid \mathrm{x}\mid +\frac{{\ln }^2\mathrm{x}}{2}+\mathrm{c}}$$

Answered by Aziztisffola last updated on 27/Jul/20

$$\mathrm{A})2\mathrm{x}\sqrt{{\mathrm{x}}^2-{\mathrm{y}}^2}{\mathrm{y}}^{\prime }=1+2\mathrm{x}\sqrt{{\mathrm{x}}^2-{\mathrm{y}}^2}$$$$(2\mathrm{x}-{({\mathrm{x}}^2-{\mathrm{y}}^2)}^{\prime })\sqrt{{\mathrm{x}}^2-{\mathrm{y}}^2}=1+2\mathrm{x}\sqrt{{\mathrm{x}}^2-{\mathrm{y}}^2}$$$$-{({\mathrm{x}}^2-{\mathrm{y}}^2)}^{\prime }\sqrt{{\mathrm{x}}^2-{\mathrm{y}}^2}=1$$$$-\frac{\mathrm{x}-{\mathrm{yy}}^{\prime }}{\sqrt{{\mathrm{x}}^2-{\mathrm{y}}^2}}\sqrt{{\mathrm{x}}^2-{\mathrm{y}}^2}=1$$$${\mathrm{yy}}^{\prime }=\mathrm{x}+1\Rightarrow \int {\mathrm{yy}}^{\prime }\mathrm{dx}=\int (\mathrm{x}+1)\mathrm{dx}$$$$\frac{{\mathrm{y}}^2}{2}=\frac{{\mathrm{x}}^2}{2}+\mathrm{x}+\mathrm{c}\Rightarrow \mathrm{y}=\underset{-}{+}\sqrt{\mid {\mathrm{x}}^2+2\mathrm{x}+\mathrm{C}\mid }$$

Answered by Aziztisffola last updated on 27/Jul/20

$$\mathrm{B})$$$$\mathrm{x}(\mathrm{y}+2){\mathrm{y}}^{\prime }=\mathrm{lnx}+1\iff (\mathrm{y}+2){\mathrm{y}}^{\prime }=\mathrm{lnx}/\mathrm{x}+1/\mathrm{x}$$$${\mathrm{yy}}^{\prime }+{2\mathrm{y}}^{\prime }=\frac{\mathrm{lnx}}{\mathrm{x}}+\frac{1}{\mathrm{x}}\Rightarrow \int ({\mathrm{yy}}^{\prime }+{2\mathrm{y}}^{\prime })\mathrm{dx}=\int (\frac{\mathrm{lnx}}{\mathrm{x}}+\frac{1}{\mathrm{x}})\mathrm{dx}$$$$\frac{1}{2}{\mathrm{y}}^2+2\mathrm{y}=\frac{1}{2}{\ln }^2\left(\mathrm{x}\right)+\mathrm{lnx}+\mathrm{C}$$$$\frac{1}{2}{\mathrm{y}}^2\left(1\right)+2\mathrm{y}\left(1\right)=\frac{1}{2}{\ln }^2\left(1\right)+\ln \left(1\right)+\mathrm{C}$$$$\frac{1}{2}-2=\mathrm{C}\Rightarrow \mathrm{C}=-\frac{3}{2}$$$$\frac{1}{2}{\mathrm{y}}^2+2\mathrm{y}=\frac{1}{2}{\ln }^2\left(\mathrm{x}\right)+\mathrm{lnx}-\frac{3}{2}$$$$\iff {\mathrm{y}}^2+4\mathrm{y}={\ln }^2\left(\mathrm{x}\right)+2\mathrm{lnx}-3$$$$\iff {\mathrm{y}}^2+4\mathrm{y}+4={\ln }^2\left(\mathrm{x}\right)+2\mathrm{lnx}+1$$$$\iff {(\mathrm{y}+2)}^2={\ln }^2\left(\mathrm{x}\right)+2\mathrm{lnx}+1$$$$\Rightarrow \mathrm{y}=\underset{-}{+}\sqrt{\mid {\ln }^2\left(\mathrm{x}\right)+2\mathrm{lnx}+1\mid }-2$$

Answered by abdomathmax last updated on 27/Jul/20

$${\mathrm{y}}^{{}^{″}}-{6\mathrm{y}}^{{}^{\prime }}-7\mathrm{y}=13\cos \left(2\mathrm{t}\right)+34\sin \left(2\mathrm{t}\right)$$$$\mathrm{h}\to {\mathrm{r}}^2-6\mathrm{r}-7=0$$$${\mathrm{\Delta }}^{{}^{\prime }}=9+7=16\Rightarrow {\mathrm{r}}_1=3+4=7\mathrm{and}{\mathrm{r}}_2=3-4=-1\Rightarrow$$$${\mathrm{y}}_{\mathrm{h}}=\mathrm{a}{\mathrm{e}}^{7\mathrm{x}}+{\mathrm{be}}^{-\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{7\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\\ {7\mathrm{e}}^{7\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right|=-{\mathrm{e}}^{6\mathrm{x}}-{7\mathrm{e}}^{6\mathrm{x}}=-{8\mathrm{e}}^{6\mathrm{x}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}{\mathrm{e}}^{-\mathrm{x}}\\ 13\cos \left(2\mathrm{x}\right)+34\sin \left(2\mathrm{x}\right)-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right|$$$$=-{\mathrm{e}}^{-\mathrm{x}}(13\cos \left(2\mathrm{x}\right)+34\sin \left(2\mathrm{x}\right))$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{e}}^{7\mathrm{x}}0\\ {7\mathrm{e}}^{7\mathrm{x}}13\cos \left(2\mathrm{x}\right)+34\sin \left(2\mathrm{x}\right)\end{array}\right|$$$$={\mathrm{e}}^{7\mathrm{x}}(13\cos \left(2\mathrm{x}\right)+34\sin \left(2\mathrm{x}\right)$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=-\int \frac{{\mathrm{e}}^{-\mathrm{x}}\{13\cos \left(2\mathrm{x}\right)+34\sin \left(2\mathrm{x}\right)\}}{-{8\mathrm{e}}^{6\mathrm{x}}}\mathrm{dx}$$$$=\frac{1}{8}\int {\mathrm{e}}^{-7\mathrm{x}}\{13\cos \left(2\mathrm{x}\right)+34\sin \left(2\mathrm{x}\right)\}\mathrm{dx}=...$$$${\mathrm{v}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=\int \frac{{\mathrm{e}}^{7\mathrm{x}}\{13\cos \left(2\mathrm{x}\right)+34\sin \left(2\mathrm{x}\right)\}}{-{8\mathrm{e}}^{6\mathrm{x}}}\mathrm{dx}$$$$=-\frac{1}{8}\int {\mathrm{e}}^{\mathrm{x}}\{13\cos \left(2\mathrm{x}\right)+34\sin \left(2\mathrm{x}\right)\}\mathrm{dx}=...$$$$\Rightarrow {\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2\mathrm{and}\mathrm{general}\mathrm{solution}\mathrm{is}$$$$\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}$$

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