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Question Number 133296 by rs4089 last updated on 21/Feb/21

Answered by SEKRET last updated on 21/Feb/21

$${\mathbf{x}}^2{\mathbf{y}}^{″}+3{\mathbf{xy}}^{\prime }+\mathbf{y}=\frac{1}{{(1-\mathbf{x})}^2}$$$${\mathbf{x}}^2{\mathbf{y}}^{″}+3{\mathbf{xy}}^{\prime }+\mathbf{y}=0\mathbf{y}={\mathbf{x}}^{\mathbf{m}}{\mathbf{y}}^{\prime }={\mathbf{mx}}^{\mathbf{m}-1}{\mathbf{y}}^{″}=\mathbf{m}(\mathbf{m}-1){\mathbf{x}}^{\mathbf{m}-2}$$$${\mathbf{x}}^2\cdot \left(\mathbf{m}(\mathbf{m}-1)\right){\mathbf{x}}^{\mathbf{m}-2}+3\mathbf{x}\cdot \mathbf{m}\cdot {\mathbf{x}}^{\mathbf{m}-1}+{\mathbf{x}}^{\mathbf{m}}=0$$$${\mathbf{m}}^2+2\mathbf{m}+1=0{\mathbf{m}}_1=-1{\mathbf{m}}_2=-1\mathbf{x}\ne 0$$$${\mathbf{y}}_1=\frac{{\mathbf{C}}_1}{\mathbf{x}}{\mathbf{y}}_2=\frac{{\mathbf{C}}_2\mathbf{ln}\left(\mathbf{x}\right)}{\mathbf{x}}\mathbf{y}={\mathbf{y}}_1+{\mathbf{y}}_2\mathbf{y}=\frac{{\mathbf{C}}_1}{\mathbf{x}}+\frac{{\mathbf{C}}_2\mathbf{ln}\left(\mathbf{x}\right)}{\mathbf{x}}$$$$\mathbf{Q}=\left(\frac{1}{\mathbf{x}}\right)\cdot \left(\frac{\mathbf{lnx}}{\mathbf{x}}\right){}^{{}^{\prime }}-\left(\frac{\mathbf{ln}\left(\mathbf{x}\right)}{\mathbf{x}}\right)\cdot {\left(\frac{1}{\mathbf{x}}\right)}^{\prime }=\frac{1}{{\mathbf{x}}^3}$$$${\mathbf{y}}^{″}+\frac{3}{\mathbf{x}}{\mathbf{y}}^{\prime }+\frac{1}{{\mathbf{x}}^2}\mathbf{y}=\frac{1}{{\mathbf{x}}^2{(1-\mathbf{x})}^2}$$$${\mathbf{u}}_1=\int \frac{-\mathbf{ln}\left(\mathbf{x}\right)}{{(1-\mathbf{x})}^2}\mathbf{dx}=\mathbf{ln}\left(\frac{\mathbf{x}}{1-\mathbf{x}}\right)+\frac{\mathbf{ln}\left(\mathbf{x}\right)}{1-\mathbf{x}}$$$${\mathbf{u}}_2=\int \frac{1}{{(\mathbf{x}-1)}^2}\mathbf{dx}=\frac{1}{1-\mathbf{x}}$$$${\mathbf{y}}_{\mathbf{u}}=\frac{1}{\mathbf{x}}\mathbf{ln}\left(\frac{\mathbf{x}}{1-\mathbf{x}}\right)$$$$\mathbf{y}=\frac{1}{\mathbf{x}}\mathbf{ln}\left(\frac{\mathbf{x}}{1-\mathbf{x}}\right)+\frac{{\mathbf{c}}_1}{\mathbf{x}}+\frac{{\mathbf{c}}_2\mathbf{ln}\left(\mathbf{x}\right)}{\mathbf{x}}$$

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