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Question Number 105163 by bemath last updated on 26/Jul/20

$$x^2{y}^{″}+{xy}^{\prime }-y=\frac{1}{1+x^2}$$

Answered by bramlex last updated on 27/Jul/20

$$wesolvehomogenousequation$$$$ThisisanEuler-Cauchy$$$$y=x^n\to \{\begin{array}{l}{y}^{\prime }={nx}^{n-1}\\ y^{″}=(n^2-n)x^{n-2}\end{array}$$$$\Rightarrow x^2(n^2-n)x^{n-2}+x\left({nx}^{n-1}\right)-x^n=0$$$$n^2-n+n-1=0\to n=\pm 1$$$$y_h=C_{1}x+\frac{C_2}{x}$$$$let{y}_1=x;y_2=x^{-1}$$$$\to \{\begin{array}{l}{y}_1^{{}^{\prime }}=1\\ y_2^{\prime }=-x^{-2}\end{array}\Rightarrow W(y_1,y_2)=\left|\begin{array}{c}x{x}^{-1}\\ 1-x^{-2}\end{array}\right|=-2x^{-1}$$$$u_1=-\int \frac{x^{-1}.\frac{1}{x^2(x^2+1)}}{-2x^{-1}}dx$$$$u_1=\int \frac{1}{2}(\frac{1}{x^2}-\frac{1}{1+x^2})dx$$$$u_1=-\frac{1}{2x}-\frac{1}{2}{\tan }^{-1}\left(x\right)$$$$u_2=\int \frac{x.\frac{1}{x^2(1+x^2)}}{-2x^{-1}}dx$$$$u_2=\int \frac{-1}{2(1+x^2)}dx=-\frac{1}{2}{\tan }^{-1}\left(x\right)$$$$particularsolution$$$$y_p=y_1{u}_1+y_2{u}_2$$$$y_p=x(-\frac{1}{2x}-\frac{1}{2}{\tan }^{-1}\left(x\right))+\frac{1}{x}(-\frac{1}{2}{\tan }^{-1}\left(x\right))$$$$y_p=-\frac{1}{2}-\frac{x}{2}{\tan }^{-1}\left(x\right)-\frac{1}{2x}{\tan }^{-1}\left(x\right)$$$$\mathcal{G}eneralsolution$$$$y_{\mathcal{G}}=C_{1}x+\frac{C_2}{x}-\frac{x}{2}{\tan }^{-1}\left(x\right)-\frac{1}{2x}{\tan }^{-1}\left(x\right)-\frac{1}{2}$$$$★\text{blacktrinagledown}◊$$

Answered by mathmax by abdo last updated on 26/Jul/20

$${\mathrm{x}}^2{\mathrm{y}}^{{}^{″}}+{\mathrm{xy}}^{{}^{\prime }}-\mathrm{y}=\frac{1}{1+{\mathrm{x}}^2}$$$$\mathrm{h}\to {\mathrm{x}}^2{\mathrm{y}}^{\left(2\right)}+{\mathrm{xy}}^{\left(1\right)}-\mathrm{y}=0\mathrm{let}\mathrm{y}={\mathrm{x}}^{\mathrm{m}}\Rightarrow {\mathrm{y}}^{{}^{\prime }}={\mathrm{mx}}^{\mathrm{m}-1}\mathrm{and}{\mathrm{y}}^{\left(2\right)}=\mathrm{m}(\mathrm{m}-1){\mathrm{x}}^{\mathrm{m}-2}$$$$\mathrm{e}\Rightarrow \mathrm{m}(\mathrm{m}-1){\mathrm{x}}^{\mathrm{m}}+{\mathrm{mx}}^{\mathrm{m}}-{\mathrm{x}}^{\mathrm{m}}=0\Rightarrow ({\mathrm{m}}^2-\mathrm{m}+\mathrm{m}-1){\mathrm{x}}^{\mathrm{m}}=0\Rightarrow$$$${\mathrm{m}}^2-1=0\Rightarrow \mathrm{m}=\stackrel{-}{+}1\Rightarrow \mathrm{y}=\alpha \mathrm{x}+\frac{\beta }{\mathrm{x}}=\alpha {\mathrm{u}}_1+\beta {\mathrm{u}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}\mathrm{x}\frac{1}{\mathrm{x}}\\ 1-\frac{1}{{\mathrm{x}}^2}\end{array}\right|=-\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}}=-\frac{2}{\mathrm{x}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}\frac{1}{\mathrm{x}}\\ \frac{1}{1+{\mathrm{x}}^2}-\frac{1}{{\mathrm{x}}^2}\end{array}\right|=-\frac{1}{\mathrm{x}(1+{\mathrm{x}}^2)}$$$${\mathrm{W}}_2=\left|\begin{array}{c}\mathrm{x}0\\ 1\frac{1}{1+{\mathrm{x}}^2}\end{array}\right|=\frac{\mathrm{x}}{1+{\mathrm{x}}^2}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=-\int \frac{1}{\mathrm{x}(1+{\mathrm{x}}^2)}\times \left(\frac{-\mathrm{x}}{2}\right)\mathrm{dx}=\frac{1}{2}\int \frac{\mathrm{dx}}{(1+{\mathrm{x}}^2)}=\frac{1}{2}\mathrm{arctanx}$$$$=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=\int \frac{\mathrm{x}}{1+{\mathrm{x}}^2}\times (-\frac{\mathrm{x}}{2})\mathrm{dx}=-\frac{1}{2}\int \frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=-\frac{1}{2}\int \frac{1+{\mathrm{x}}^2-1}{1+{\mathrm{x}}^2}\mathrm{dx}=-\frac{\mathrm{x}}{2}+\frac{1}{2}\mathrm{arctanx}\Rightarrow$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2=\frac{\mathrm{x}}{2}\mathrm{arctanx}+\frac{1}{\mathrm{x}}(-\frac{\mathrm{x}}{2}+\frac{1}{2}\mathrm{arctanx})$$$$=\frac{\mathrm{x}}{2}\mathrm{arctanx}-\frac{1}{2}+\frac{1}{2\mathrm{x}}\mathrm{arctanx}=\frac{1}{2}(\mathrm{x}+\frac{1}{\mathrm{x}})\mathrm{arctanx}-\frac{1}{2}$$$$\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}$$$$\mathrm{y}=\alpha \mathrm{x}+\beta {\mathrm{x}}^{-1}+\frac{1}{2}(\mathrm{x}+\frac{1}{\mathrm{x}})\mathrm{arctanx}-\frac{1}{2}$$$$$$

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