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Question Number 99260 by Ar Brandon last updated on 19/Jun/20

$$\mathrm{Solve}\mathrm{the}\mathrm{differential}\mathrm{equation}$$$${\mathrm{xy}}^{\prime }-\mathrm{y}+\frac{2\mathrm{x}+1}{{(\mathrm{x}+1)}^2}=0$$

Answered by Mr.D.N. last updated on 19/Jun/20

Commented by Ar Brandon last updated on 20/Jun/20

Thank you ��

Answered by mathmax by abdo last updated on 19/Jun/20

$${\mathrm{xy}}^{{}^{\prime }}-\mathrm{y}+\frac{2\mathrm{x}+1}{{(\mathrm{x}+1)}^2}=0\left(\mathrm{he}\right)\to {\mathrm{xy}}^{{}^{\prime }}-\mathrm{y}=0\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\frac{1}{\mathrm{x}}\Rightarrow \ln \mid \mathrm{y}\mid =\ln \mid \mathrm{x}\mid +\mathrm{c}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{k}\mid \mathrm{x}\mid \mathrm{let}\mathrm{determine}\mathrm{slution}\mathrm{on}]0,+\mathrm{\infty }[\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=\mathrm{kx}$$$$\mathrm{mvc}\mathrm{method}\to {\mathrm{y}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}\mathrm{x}+\mathrm{k}$$$$\left(\mathrm{e}\right)\Rightarrow {\mathrm{k}}^{{}^{\prime }}{\mathrm{x}}^2+\mathrm{kx}-\mathrm{kx}=-\frac{2\mathrm{x}+1}{{(\mathrm{x}+1)}^2}\Rightarrow {\mathrm{k}}^{{}^{\prime }}=-\frac{2\mathrm{x}+1}{{\mathrm{x}}^2{(\mathrm{x}+1)}^2}\Rightarrow \mathrm{k}\left(\mathrm{x}\right)=-\int \frac{2\mathrm{x}+1}{{\mathrm{x}}^2{(\mathrm{x}+1)}^2}\mathrm{dx}+\mathrm{c}$$$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{2\mathrm{x}+1}{{\mathrm{x}}^2{(\mathrm{x}+1)}^2}\mathrm{we}\mathrm{have}\frac{1}{{\mathrm{x}}^2}-\frac{1}{{(\mathrm{x}+1)}^2}=\frac{{(\mathrm{x}+1)}^2-{\mathrm{x}}^2}{{\mathrm{x}}^2{(\mathrm{x}+1)}^2}$$$$=\frac{2\mathrm{x}+1}{{\mathrm{x}}^2{(\mathrm{x}+1)}^2}=\mathrm{F}\left(\mathrm{x}\right)\Rightarrow \mathrm{k}\left(\mathrm{x}\right)=-\int \frac{\mathrm{dx}}{{\mathrm{x}}^2}+\int \frac{\mathrm{dx}}{{(\mathrm{x}+1)}^2}=\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1}+\mathrm{c}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{xk}\left(\mathrm{x}\right)=\mathrm{x}(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1}+\mathrm{c})=1-\frac{\mathrm{x}}{\mathrm{x}+1}+\mathrm{c}=\frac{1}{\mathrm{x}+1}+\mathrm{c}$$

Commented by mathmax by abdo last updated on 19/Jun/20

$$\mathrm{sorry}\mathrm{y}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}+1}+\mathrm{cx}$$

Commented by Ar Brandon last updated on 20/Jun/20

Thanks

Commented by mathmax by abdo last updated on 20/Jun/20

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

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