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Question Number 108475 by pticantor last updated on 17/Aug/20

$$(x^2+1){\mathit{y}}^{{}^{\prime }}+2x\mathit{y}=x\sqrt{x^2+1}$$$$pleasesolvethisdifferential$$$$equation$$

Answered by Dwaipayan Shikari last updated on 17/Aug/20

$$\frac{dy}{dx}+\frac{2x}{x^2+1}y=\frac{x}{\sqrt{x^2+1}}$$$$I.F=e^{\int \frac{2x}{x^2+1}}=e^{log(x^2+1)}=x^2+1$$$$y.(x^2+1)=\int \frac{x^2+1}{\sqrt{x^2+1}}.x$$$$y(x^2+1)=\frac{1}{2}\int 2x.\sqrt{x^2+1}dx$$$$y(x^2+1)=\frac{1}{3}{(x^2+1)}^{\frac{3}{2}}+C$$$$y=\frac{1}{3}\sqrt{x^2+1}+\frac{C}{(x^2+1)}$$

Answered by mathmax by abdo last updated on 17/Aug/20

$$({\mathrm{x}}^2+1){\mathrm{y}}^{{}^{\prime }}+2\mathrm{xy}=\mathrm{x}\sqrt{{\mathrm{x}}^2+1}$$$$\mathrm{h}\to ({\mathrm{x}}^2+1){\mathrm{y}}^{{}^{\prime }}=-2\mathrm{xy}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\frac{-2\mathrm{x}}{{\mathrm{x}}^2+1}\Rightarrow \ln \mid \mathrm{y}\mid =-\ln ({\mathrm{x}}^2+1)+\mathrm{c}\Rightarrow$$$$\mathrm{y}=\frac{\mathrm{k}}{{\mathrm{x}}^2+1}\mathrm{lagrange}\mathrm{method}\Rightarrow {\mathrm{y}}^{{}^{\prime }}=\frac{{\mathrm{k}}^{{}^{\prime }}}{{\mathrm{x}}^2+1}-\frac{2\mathrm{xk}}{{({\mathrm{x}}^2+1)}^2}$$$$\mathrm{e}\Rightarrow {\mathrm{k}}^{{}^{\prime }}-\frac{2\mathrm{xk}}{{\mathrm{x}}^2+1}+\frac{2\mathrm{xk}}{{\mathrm{x}}^2+1}=\mathrm{x}\sqrt{{\mathrm{x}}^2+1}\Rightarrow {\mathrm{k}}^{{}^{\prime }}=\mathrm{x}\sqrt{{\mathrm{x}}^2+1}\Rightarrow$$$$\mathrm{k}\left(\mathrm{x}\right)=\int \mathrm{x}\sqrt{1+{\mathrm{x}}^2}\mathrm{dx}{=}_{\mathrm{x}=\mathrm{sht}}\int \mathrm{sht}{\mathrm{c}}^2\mathrm{ht}\mathrm{dt}=\frac{1}{3}{\mathrm{ch}}^3\mathrm{t}+\mathrm{c}$$$$=\frac{1}{3}{\left(\frac{{\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}}}{2}\right)}^3+\mathrm{c}=\frac{1}{24}{({\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}})}^3+\mathrm{c}\mathrm{we}\mathrm{have}\mathrm{t}=\ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})\Rightarrow$$$$\mathrm{k}\left(\mathrm{x}\right)=\frac{1}{24}{\{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}+\frac{1}{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}\}}^3+\mathrm{c}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^2+1}\{\frac{1}{24}{(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}+\frac{1}{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}})}^3+\mathrm{c}\}$$

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