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Question Number 50432 by Abdo msup. last updated on 16/Dec/18

$$solvex(x^2-1)y^{{}^{\prime }}+2y=x^2$$

Commented by Abdo msup. last updated on 16/Dec/18

$$\left(he\right)\Rightarrow x(x^2-1)y^{{}^{\prime }}+2y=0\iff x(x^2-1)y^{{}^{\prime }}=-2y\iff$$$$\frac{y^{{}^{\prime }}}{y}=\frac{-2}{x(x^2-1)}\Rightarrow ln\mid y\mid =-2\int \frac{dx}{x(x^2-1)}+cletdecompise$$$$F\left(x\right)=\frac{1}{x(x^2-1)}\Rightarrow F\left(x\right)=\frac{1}{x(x-1)(x+1)}=\frac{a}{x}+\frac{b}{x-1}+\frac{c}{x+1}$$$$a={lim}_{x\to o}xF\left(x\right)=-1$$$$b={lim}_{x\to 1}(x-1)F\left(x\right)=\frac{1}{2}$$$$c={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{1}{2}\Rightarrow F\left(x\right)=\frac{-1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}$$$$\Rightarrow ln\mid y\mid =-2\int (\frac{-1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)})dx$$$$=2ln\mid x\mid -ln\mid x-1\mid -ln\mid x+1\mid +c\Rightarrow$$$$y\left(x\right)=k\frac{x^2}{\mid x^2-1\mid }letsuppose\mid x\mid >1\Rightarrow$$$$y\left(x\right)=K\frac{x^2}{x^2-1}mvcmethodgive$$$$y^{{}^{\prime }}=K^{{}^{\prime }}\frac{x^2}{x^2-1}+K\frac{2x(x^2-1)-x^2\left(2x\right)}{{(x^2-1)}^2}$$$$=K^{{}^{\prime }}\frac{x^2}{x^2-1}+K\frac{-2x}{{(x^2-1)}^2}and\left(e\right)\iff$$$$x(x^2-1)(K^{{}^{\prime }}\frac{x^2}{x^2-1}+K\frac{-2x}{{(x^2-1)}^2})+2K\frac{x^2}{x^2-1}=x^2\Rightarrow$$$$K^{{}^{\prime }}{x}^3-2x^{2}K\frac{1}{x^2-1}+2K\frac{x^2}{x^2-1}=x^2\Rightarrow K^{{}^{\prime }}=\frac{1}{x}\Rightarrow$$$$K=ln\mid x\mid +\lambda \Rightarrow$$$$y\left(x\right)=\frac{x^2}{x^2-1}(ln\mid x\mid +\lambda )=\frac{x^{2}ln\mid x\mid }{x^2-1}+\frac{\lambda x^2}{x^2-1}.$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Dec/18

$$\frac{dy}{dx}+\frac{2}{x(x^2-1)}y=\frac{x}{x^2-1}$$$$I.F{e}^{\int \frac{2}{x(x^2-1)}dx}$$$$\int \frac{2x}{x^2(x^2-1)}dx$$$$\int \frac{dt}{t(t-1)}=\int \frac{t-(t-1)}{t(t-1)}dt=\int \frac{dt}{t-1}-\int \frac{dt}{t}=ln\left(\frac{t-1}{t}\right)$$$$I.F{e}^{ln\left(\frac{x^2-1}{x^2}\right)}=\frac{x^2-1}{x^2}$$$$\frac{x^2-1}{x^2}\frac{dy}{dx}+\frac{x^2-1}{x^2}\times \frac{2}{x(x^2-1)}y=\frac{x^2-1}{x^2}\times \frac{x}{x^2-1}$$$$(1-\frac{1}{x^2})\frac{dy}{dx}+\left(\frac{2}{x^3}\right)y=\frac{1}{x}$$$$\frac{d}{dx}\left\{y\times (1-\frac{1}{x^2})\right\}=\frac{1}{x}$$$$d\left\{y\times (1-\frac{1}{x^2})\right\}=\frac{dx}{x}$$$$intregating$$$$y\times (1-\frac{1}{x^2})=lnx+c$$$$$$$$$$$$$$

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