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Question Number 140896 by bramlexs22 last updated on 14/May/21

$$\mathrm{the}\mathrm{function}\mathrm{f}\mathrm{with}\mathrm{variable}\mathrm{x}$$$$\mathrm{satisfies}\mathrm{the}\mathrm{equation}$$$${\mathrm{x}}^2\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)+2\mathrm{x}\mathrm{f}\left(\mathrm{x}\right)=\arctan \mathrm{x}\mathrm{for}$$$$0<\arctan \mathrm{x}<\frac{\pi }{2}\mathrm{and}\mathrm{f}\left(1\right)=\frac{\pi }{4}.$$$$\mathrm{find}\mathrm{f}\left(\mathrm{x}\right).$$

Answered by EDWIN88 last updated on 15/May/21

$$\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right),\mathrm{diff}\mathrm{wrt}\mathrm{x}\mathrm{give}$$$$\mathrm{g}{}^{\prime }\left(\mathrm{x}\right)={\mathrm{x}}^2\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)+2\mathrm{x}\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{so}\mathrm{that}\mathrm{g}{}^{\prime }\left(\mathrm{x}\right)=\arctan \mathrm{x}$$$$\mathrm{integration}\mathrm{by}\mathrm{part}$$$$\mathrm{g}\left(\mathrm{x}\right)=\int \arctan \mathrm{x}\mathrm{dx}=\mathrm{x}\arctan \mathrm{x}-\frac{1}{2}\ln (1+{\mathrm{x}}^2)+\mathrm{c}$$$$\mathrm{the}\mathrm{condition}\mathrm{g}\left(1\right)=\mathrm{f}\left(1\right)=\frac{\pi }{4}$$$$\Rightarrow \cancel{\frac{\pi }{4}}=\cancel{\arctan 1}-\frac{1}{2}\ln \left(2\right)+\mathrm{c}$$$$\Rightarrow \mathrm{c}=\ln \sqrt{2}$$$$\therefore \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{g}\left(\mathrm{x}\right)}{{\mathrm{x}}^2}=\frac{\arctan \mathrm{x}}{\mathrm{x}}-\frac{\ln \sqrt{1+{\mathrm{x}}^2}}{{\mathrm{x}}^2}+\frac{\ln \sqrt{2}}{{\mathrm{x}}^2}$$

Answered by mathmax by abdo last updated on 14/May/21

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{y}\mathrm{e}\Rightarrow {\mathrm{x}}^2{\mathrm{y}}^{{}^{\prime }}+2\mathrm{xy}=\mathrm{arctanx}\mathrm{and}\mathrm{y}\left(1\right)=\frac{\pi }{4}$$$$\mathrm{h}\to {\mathrm{xy}}^{{}^{\prime }}+2\mathrm{y}=0\Rightarrow {\mathrm{xy}}^{{}^{\prime }}=-2\mathrm{y}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=-\frac{2}{\mathrm{x}}\Rightarrow \ln \mid \mathrm{y}\mid =-2\log \mid \mathrm{x}\mid +\mathrm{c}\Rightarrow$$$$\mathrm{y}=\frac{\mathrm{k}}{{\mathrm{x}}^2}\mathrm{mvc}\mathrm{method}\to {\mathrm{y}}^{{}^{\prime }}=\frac{{\mathrm{k}}^{{}^{\prime }}}{{\mathrm{x}}^2}+\mathrm{k}\left(\frac{-2\mathrm{x}}{{\mathrm{x}}^4}\right)=\frac{{\mathrm{k}}^{{}^{\prime }}}{{\mathrm{x}}^2}-\frac{2\mathrm{k}}{{\mathrm{x}}^3}$$$$\mathrm{e}\Rightarrow {\mathrm{k}}^{{}^{\prime }}-\frac{2\mathrm{k}}{\mathrm{x}}+\frac{2\mathrm{k}}{\mathrm{x}}=\mathrm{arctanx}\Rightarrow \mathrm{k}=\int \mathrm{arctanx}\mathrm{dx}$$$$=\mathrm{xarctanx}-\int \frac{\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}=\mathrm{xarctan}\left(\mathrm{x}\right)-\frac{1}{2}\log (1+{\mathrm{x}}^2)+\mathrm{C}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^2}(\mathrm{xarctan}\left(\mathrm{x}\right)-\frac{1}{2}\log (1+{\mathrm{x}}^2)+\mathrm{C})$$$$=\frac{\arctan \left(\mathrm{x}\right)}{\mathrm{x}}-\frac{1}{{2\mathrm{x}}^2}\log (1+{\mathrm{x}}^2)+\frac{\mathrm{c}}{{\mathrm{x}}^2}=\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{f}\left(1\right)=\frac{\pi }{4}\Rightarrow \frac{\pi }{4}-\frac{1}{2}\log 2+\mathrm{C}=\frac{\pi }{4}\Rightarrow \mathrm{C}=\frac{1}{2}\log 2\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\arctan \left(\mathrm{x}\right)}{\mathrm{x}}-\frac{1}{{2\mathrm{x}}^2}\log (1+{\mathrm{x}}^2)+\frac{\log 2}{{2\mathrm{x}}^2}$$

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