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Please help me find two independent solutions for $$3x(x+1)y''+2(x+1)y'+4y=0$$ Thanks from a new beginner into ODE's.

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See this link en.wikipedia.org/wiki/Regular_singular_point. –  Babak S. Jan 3 '13 at 17:28
    
Use Frobenius method. Note that, the ode has a solution in terms of the hypergeometric function. –  Mhenni Benghorbal Jan 3 '13 at 17:29

3 Answers 3

up vote 1 down vote accepted

We suppose that the $\displaystyle y=\sum_{n=0}^{\infty}a_{n}x^n$ is a solution.

Then

$$y=a_{0}+a_{1}x+...+a_{n}x^n+... $$

$$y'=a_{1}+2a_{2}x+3a_{3}x^2+4a_{4}x^3+...+(n+1)a_{n+1}x^n+...$$

$$ y''=2a_{2}+3.2a_{3}x+4.3a_{4}x^2+...+(n+2)(n+1)a_{n+2}x^n+... $$

replacing this in the DE,

$3x^2y''+3xy''+2xy'+2y'+4y=0$,

we have

general term
$$3(n+2)(n+1)a_{n+2}x^{n+2}+3(n+3)(n+2)a_{n+3}x^{n+2}+2(n+2)a_{n+2}x^{n+2}+2(n+3)a_{n+3}x^{n+2}+4a_{n+2}x^{n+2}$$

Then we must have

$$3(n+2)(n+1)a_{n+2}+3(n+3)(n+2)a_{n+3}+2(n+2)a_{n+2}+2(n+3)a_{n+3}+4a_{n+2}=0$$ or

$$[3(n+3)(n+2)+2(n+3)]a_{n+3}+[3(n+2)(n+1)+2(n+2)+4]a_{n+2}=0$$

Now you solve this difference equation of the first order to $a_{0}=0$,$a_{1}=1$ and $a_{0}=1, a_{1}=0$. This give you two solution LI.

Note that $a_{0}$ and $a_{1}$ are the initial condition $y(0)$ and $y'(0)$

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Note that both $x=0, x=1$ are regular singular points( check them!). As @Edgar assumed, let $y=\sum_1^{\infty}a_nx^n$ and by writing the equation like $$x^2y''+\frac{2}{3}xy'+\frac{4x}{3(x+1)}y=0$$ we get: $$p(x)=\frac{2}{3},\; q(x)=\frac{4x}{3(x+1)}$$ and then $p(0)=\frac{2}{3},\; q(0)=0$. Now I suggest you to set the characteristic equation (see http://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdf) $$s(s-1)+sp(0)+q(0)=0$$ You get $s_1=0,s_2=\frac{1}{3}$. Can you do the rest from here?

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Since $x=0$ is a (regular) singular point of your ODE, if you want a power series solution about $x=0$, you will need to appeal to the Method of Frobenius to construct it.

This is a lengthy process, but not overly hard if you work through the tedium.

Here are some intermediate steps to help you check your work along the way:

  • The indicial equation is $r(r-1)+{2\over 3}r=0$ so the roots are $r_1=1/3$ and $r_2=0$.

  • Since $r_1-r_2\not\in\mathbb{Z}$, the situation is easier than it would have been otherwise. Two linearly independent solutions are \begin{align} y_1&=|x|^{r_1}\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_nx^{n+1/3},\\ y_2&=|x|^{r_2}\sum_{n=0}^\infty b_nx^n=\sum_{n=0}^\infty b_nx^n. \end{align}

  • To determine the coefficients for each, plug each series back into the ODE and group the coefficients by like powers to obtain a recursion formula for the $n$th coefficient. For a detailed example, see this particular example from the link above.

Hope that helps.

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