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Question Number 184858 by SANOGO last updated on 12/Jan/23

$$\underset{n=o}{\sum ^{+oo}}\frac{{(-1)}^n{x}^{2n+1}}{4n^2-1}$$

Answered by qaz last updated on 14/Jan/23

$$y=\mathrm{\Sigma }\frac{{(-1)}^n}{4n^2-1}{x}^{2n+1}$$$$y^{\prime }=\mathrm{\Sigma }\frac{{(-1)}^n{x}^{2n}}{2n-1}=x\mathrm{\Sigma }\frac{{(-1)}^n{x}^{2n-1}}{2n-1}$$$$y^{″}=\mathrm{\Sigma }\frac{{(-1)}^n{x}^{2n-1}}{2n-1}+x\mathrm{\Sigma }{(-1)}^n{x}^{2n-2}=\frac{y^{\prime }}{x}+\frac{1}{x(1+x^2)}$$$$\Rightarrow y^{\prime }=C_{1}x-1-x\arctan x$$$$\Rightarrow y=C_1{x}^2-\frac{1}{2}x-\frac{1}{2}{x}^2\arctan x-\frac{1}{2}\arctan x+C_2$$$$y\left(0\right)=0y{\left(0\right)}^{\prime }=-1\Rightarrow C_1=0C_2=0$$$$\Rightarrow y=-\frac{1}{2}x-\frac{1}{2}{x}^2\arctan x-\frac{1}{2}\arctan x$$

Commented by SANOGO last updated on 14/Jan/23

$$thankyou$$

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