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$$y''(x)+\left(\dfrac 2 x +\dfrac x 2\right)y'(x)+ \dfrac 3 2 y(x)=0$$

Can anyone help me by telling me a method to solve this kind of differential equations ? I tried Cauchy Euler method, but it did not work.

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The term $\frac 2x+\frac x2=\frac{x^2+4}{2x}$ gave me the idea to define $y=e^{-\frac{x^2}{4}} z$ which makes the equation $$e^{-\frac{x^2}{4}}\frac{ \left(4-x^2\right) z'(x)+2 x z''(x)}{2 x}=0$$ So, consider $$\left(4-x^2\right) z'+2 x z''=0$$ Now, reduce the order $u=z'$ to get $$\left(4-x^2\right) u+2 x u'=0$$ which is separable and leads to $$u=z'=\frac{c_1}{x^2} e^{\frac{x^2}{4}}\implies z=c_1 \left(\frac{1}{2} \sqrt{\pi } \text{erfi}\left(\frac{x}{2}\right)-\frac{e^{\frac{x^2}{4}}}{x}\right)+c_2$$

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  • $\begingroup$ Thanks. what is erfi ? $\endgroup$ – Mike Apr 30 '18 at 3:54
  • $\begingroup$ @Mike. The imaginary error function. Have a look at mathworld.wolfram.com/Erfi.html $\endgroup$ – Claude Leibovici Apr 30 '18 at 4:27
  • $\begingroup$ +1 Claude for your nice solution $\endgroup$ – Isham Apr 30 '18 at 4:36
  • $\begingroup$ @ClaudeLeibovici I really appreciate your help. There was given two conditions that I do not understand how to apply them. First one, limit of y(x)= 0 when x goes to infinity. Second one, the integral from 0 to infinity of square x multiply by y(x)= k/4pi. Please help me $\endgroup$ – Mike Apr 30 '18 at 5:08
  • $\begingroup$ @Mike Please update the question to include all the conditions $\endgroup$ – Dylan Apr 30 '18 at 12:41
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$$y''(x)+(\dfrac 2x +\dfrac x2)y'(x)+ (\dfrac32) y(x)=0$$ Multiply by $2x^2$ $$(2x^2y')'+(x^3y)'=0$$ Integrate $$2x^2y'+x^3y=K$$ $$y'+\frac x2y=\frac K {2x^2}$$ $$y'+\frac x2y=\frac C{x^2}$$ You can try to solve the first ode ( It may not be expressed with elementary functions) $$(ye^{x^2/4})'=e^{x^2/4}\frac C{x^2}$$ $$y=C_1e^{-x^2/4} \int \frac {e^{x^2/4}}{x^2}dx+C_2e^{-x^2/4}$$

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    $\begingroup$ Nice manipulation ! $\to +1$ $\endgroup$ – Claude Leibovici Apr 30 '18 at 4:33
  • $\begingroup$ Thanks @ClaudeLeibovici yours is also nice and more complete .. $\endgroup$ – Isham Apr 30 '18 at 4:35
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    $\begingroup$ thank you all of you.. I will consider that next time. $\endgroup$ – Mike May 2 '18 at 22:28
  • $\begingroup$ yw @Mike......... $\endgroup$ – Isham May 2 '18 at 22:35

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