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Question Number 143878 by akmalovna05 last updated on 19/Jun/21

$${\mathrm{y}}^{‴}={2\mathrm{xy}}^{″}$$

Answered by Ar Brandon last updated on 19/Jun/21

$${\mathrm{y}}^{‴}={2\mathrm{xy}}^{″}\Rightarrow \frac{{\mathrm{y}}^{‴}}{{\mathrm{y}}^{″}}=2\mathrm{x}\Rightarrow \ln \left({\mathrm{y}}^{″}\right)={\mathrm{x}}^2\Rightarrow {\mathrm{y}}^{″}={\mathrm{e}}^{{\mathrm{x}}^2}$$

Answered by ajfour last updated on 19/Jun/21

$$d\left(\frac{d^{2}y}{{dx}^2}\right)=2x\left(\frac{d^{2}y}{{dx}^2}\right)dx$$$$\Rightarrow \ln \left(\frac{d^{2}y}{{dx}^2}\right)=x^2+c$$$$d\left(\frac{dy}{dx}\right)={ke}^{x^2}$$$$\frac{dy}{dx}=(k\int e^{x^2}dx)+b$$$$y=\int (k\int e^{x^2}dx)dx+bx+a$$

Answered by Olaf_Thorendsen last updated on 19/Jun/21

$$y^{‴}=2{xy}^{″}$$$$\frac{y^{‴}}{y^{″}}=2x$$$$\ln \mid y^{″}\mid =x^2+{\mathrm{C}}_1$$$$y^{″}={\mathrm{C}}_2{e}^{x^2}$$$$y^{\prime }=\int {\mathrm{C}}_2{e}^{x^2}dx={\mathrm{C}}_2\left(\frac{\sqrt{\pi }}{2}\mathrm{erfi}\left(x\right)\right)+{\mathrm{C}}_3$$$$y^{\prime }={\mathrm{C}}_4\mathrm{erfi}\left(x\right)+{\mathrm{C}}_3$$$$\mathrm{erfi}\left(x\right)=\frac{2}{\sqrt{\pi }}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(2n+1)n!}{x}^{2n+1}$$$$y^{\prime }={\mathrm{C}}_5\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(2n+1)n!}{x}^{2n+1}+{\mathrm{C}}_3$$$$y={\mathrm{C}}_5\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(2n+1)(2n+2)n!}{x}^{2n+2}+{\mathrm{C}}_{3}x+{\mathrm{C}}_6$$$$y=\frac{{\mathrm{C}}_5}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+1)(2n+1)n!}{x}^{2n+2}+{\mathrm{C}}_{3}x+{\mathrm{C}}_6$$$$\mathrm{Finally}:$$$$y=a\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+1)(2n+1)n!}{x}^{2n+2}+bx+c$$

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