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Thank you so much in advance for any help that you might give me. I'm stuck and can't solve the following equation

$y(x)+C_1y'(x) x+C_2 y''(x)x²=0$

Where $C_1$ and $C_2$ are constants.

More importantly, $-r y(x)+\theta (\mu - x)y'(x)+\frac{\sigma ²}{2}y''(x)x²=0$ Any help?

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  • $\begingroup$ Does $\theta$ denote the Heaviside function? $\endgroup$ – mickep Jan 14 '16 at 15:38
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    $\begingroup$ Are you sure that the first term in the first DE is $y'(x)$ ? Could it be $y(x)$ instead ? $\endgroup$ – Claude Leibovici Jan 14 '16 at 15:40
  • $\begingroup$ i'm sure it must be $$y(x)$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 14 '16 at 15:41
  • $\begingroup$ Is there a $x^2$ missing in the 2nd eq? $\endgroup$ – Elsa Jan 14 '16 at 16:17
  • $\begingroup$ For the first one, if $y$ is smooth, use the Maclaurin series of $y(x)$. $\endgroup$ – Hetebrij Jan 14 '16 at 21:49
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Looking at the first equation:

$0 =y(x)+cy'(x) x+d y''(x)x^2 $.

Let $y(x) =\sum_{n=0}^{\infty} a_nx^n $.

$y'(x) =\sum_{n=1}^{\infty} na_nx^{n-1} $ so $cxy'(x) =\sum_{n=1}^{\infty} cna_nx^{n} $.

$y''(x) =\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} $ so $dx^2y''(x) =\sum_{n=2}^{\infty} dn(n-1)a_nx^{n} $.

Therefore

$\begin{array}\\ 0 &=y(x)+cy'(x) x+d y''(x)x^2\\ &=a_0+a_1x+ca_1x +\sum_{n=2}^{\infty}a_nx^n(1+cn+dn(n-1))\\ \end{array} $

Therefore, no solution of this type.

If $y = ax^b$, $cxy' = abcx^{b}$ and $dx^2y'' = ab(b-1)dx^{b}$, so $0 =x^b(a+abc+ab(b-1)d) =ax^b(1+bc+b(b-1)d) $ so

$\begin{array}\\ 0 &=1+bc+b(b-1)d\\ &=1+bc-bd+b^2d\\ &=1+b(c-d)+b^2d\\ \end{array} $

Therefore, if $d \ne 0$,

$\begin{array}\\ b &=\frac{d-c\pm\sqrt{(c-d)^2-4d}}{2d}\\ &=\frac{d-c\pm\sqrt{c^2-6cd+d^2}}{2d}\\ &=\frac{1-r\pm\sqrt{r^2-6r+1}}{2} \qquad\text{where }r = \frac{c}{d}\\ \end{array} $

If $d = 0$, $b = -\frac1{c} $.

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    $\begingroup$ If there is a $n$ such that $1+cn+dn(n-1)=0$, the series give the solution $y=a x^n$. $\endgroup$ – Hetebrij Jan 14 '16 at 23:21

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