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Question Number 132432 by Chhing last updated on 14/Feb/21

$$$$$$\mathrm{Solve}\mathrm{differential}\mathrm{equations}$$$$1/({\mathrm{x}}^3-1){\mathrm{y}}^{\prime }+\mathrm{xy}=\mathrm{x}$$$$2/({\mathrm{x}}^2-1){\mathrm{y}}^{\prime }-\mathrm{xy}+\frac{2\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}=0$$$$3/{\mathrm{x}}^2\left(\mathrm{lnx}\right){\mathrm{y}}^{″}+\mathrm{y}=0,\mathrm{know}\mathrm{that}\mathrm{y}=\mathrm{lnx}\mathrm{is}\mathrm{the}\mathrm{answer}$$$$$$

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

$$1)\mathrm{h}\to ({\mathrm{x}}^3-1){\mathrm{y}}^{{}^{\prime }}+\mathrm{xy}=0\Rightarrow ({\mathrm{x}}^3-1){\mathrm{y}}^{{}^{\prime }}=-\mathrm{xy}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\frac{-\mathrm{x}}{{\mathrm{x}}^3-1}\Rightarrow$$$$\ln \mid \mathrm{y}\mid =-\int \frac{\mathrm{xdx}}{{\mathrm{x}}^3-1}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\mathrm{x}}^3-1}=\frac{\mathrm{x}}{(\mathrm{x}-1)({\mathrm{x}}^2+\mathrm{x}+1}$$$$=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{\mathrm{bx}+\mathrm{c}}{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{we}\mathrm{have}\mathrm{a}=\frac{1}{3}$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{xF}\left(\mathrm{x}\right)=0=\mathrm{a}+\mathrm{b}\Rightarrow \mathrm{b}=-\frac{1}{3}$$$$\mathrm{F}\left(0\right)=0=-\mathrm{a}+\mathrm{c}\Rightarrow \mathrm{c}=\mathrm{a}=\frac{1}{3}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{1}{3(\mathrm{x}-1)}+\frac{-\frac{1}{3}\mathrm{x}+\frac{1}{3}}{{\mathrm{x}}^2+\mathrm{x}+1}$$$$\int \mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=\frac{1}{3}\ln \mid \mathrm{x}-1\mid -\frac{1}{3}\int \frac{\mathrm{x}-1}{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}$$$$=\frac{1}{3}\ln \mid \mathrm{x}-1\mid -\frac{1}{6}\int \frac{2\mathrm{x}+1-3}{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}$$$$=\frac{1}{3}\ln \mid \mathrm{x}-1\mid -\frac{1}{6}\ln ({\mathrm{x}}^2+\mathrm{x}+1)+\frac{1}{2}\int \frac{\mathrm{dx}}{{(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4}}$$$$\int \frac{\mathrm{dx}}{{(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4}}{=}_{\mathrm{x}+\frac{1}{2}=\frac{\sqrt{3}}{2}\mathrm{t}}\frac{4}{3}\int \frac{1}{{\mathrm{t}}^2+1}\frac{\sqrt{3}}{2}\mathrm{dt}$$$$=\frac{2}{\sqrt{3}}\arctan \left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)+\mathrm{c}\Rightarrow$$$$\ln \mid \mathrm{y}\mid =\frac{1}{3}\ln \mid \mathrm{x}-1\mid -\frac{1}{6}\ln ({\mathrm{x}}^2+\mathrm{x}+1)+\frac{1}{\sqrt{3}}\arctan \left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)+\mathrm{c}\Rightarrow$$$$\mathrm{y}=\mathrm{k}\mid \mathrm{x}-1{\mid }^3.{({\mathrm{x}}^2+\mathrm{x}+1)}^{-\frac{1}{6}}.{\mathrm{e}}^{\frac{1}{\sqrt{3}}\arctan \left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)}$$$$....\mathrm{be}\mathrm{continued}....$$

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

$$2)({\mathrm{x}}^2-1){\mathrm{y}}^{{}^{\prime }}-\mathrm{xy}+\frac{2\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}=0$$$$\mathrm{h}\to ({\mathrm{x}}^2-1){\mathrm{y}}^{{}^{\prime }}=\mathrm{xy}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\frac{\mathrm{x}}{{\mathrm{x}}^2-1}\Rightarrow \ln \mid \mathrm{y}\mid =\int \frac{\mathrm{xdx}}{{\mathrm{x}}^2-1}+\mathrm{c}$$$$=\frac{1}{2}\ln \mid {\mathrm{x}}^2-1\mid +\mathrm{c}\Rightarrow \mathrm{y}=\mathrm{k}\sqrt{\mid {\mathrm{x}}^2-1\mid }\mathrm{solution}\mathrm{on}\mathrm{w}=\{\mathrm{x}/{\mathrm{x}}^2-1>0\}$$$$\Rightarrow \mathrm{y}=\mathrm{k}\sqrt{{\mathrm{x}}^2-1}\mathrm{lagrange}\mathrm{method}{\mathrm{y}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}\sqrt{{\mathrm{x}}^2-1}+\mathrm{k}\frac{2\mathrm{x}}{2\sqrt{{\mathrm{x}}^2-1}}$$$$\mathrm{e}\Rightarrow ({\mathrm{x}}^2-1){\mathrm{k}}^{{}^{\prime }}\sqrt{{\mathrm{x}}^2-1}+\mathrm{kx}\sqrt{{\mathrm{x}}^2-1}-\mathrm{xk}\sqrt{{\mathrm{x}}^2-1}+\frac{2\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}=0\Rightarrow$$$${\mathrm{k}}^{{}^{\prime }}({\mathrm{x}}^2-1)\sqrt{{\mathrm{x}}^2-1}=-\frac{2\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}\Rightarrow {\mathrm{k}}^{{}^{\prime }}=-\frac{2\mathrm{x}}{({\mathrm{x}}^2-1)\sqrt{{\mathrm{x}}^2-1}\sqrt{{\mathrm{x}}^2+1}}\Rightarrow$$$${\mathrm{k}}^{{}^{\prime }}=\frac{2\mathrm{x}}{({\mathrm{x}}^2-1)\sqrt{{\mathrm{x}}^4-1}}\Rightarrow \mathrm{k}\left(\mathrm{x}\right)=\int \frac{2\mathrm{xdx}}{({\mathrm{x}}^2-1)\sqrt{{\mathrm{x}}^4-1}}+\lambda \Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\sqrt{{\mathrm{x}}^2-1}\{\int \frac{2\mathrm{x}}{({\mathrm{x}}^2-1)\sqrt{{\mathrm{x}}^4-1}}\mathrm{dx}+\lambda \}$$

Commented by Chhing last updated on 15/Feb/21

$$\mathrm{Thank}\mathrm{sir}$$

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