I am a tad confused about what is going on. The process is:

  • assume y can be represented by a power series
  • find the derivatives of y and plug them into the differential equation
  • try to find the coefficient of the power series through some kind of recursive formula

Is that right?

Example. Can you check my work?

  1. $$(x-3)y' + 2y = 0$$

$$y = \sum_{n=0}^\infty c_nx^n$$ $$y' = \sum_{n=1}^\infty n\cdot c_nx^{n -1} = \sum_{n=0}^\infty n\cdot c_nx^{n -1}$$ the $y'$ terms are equal because the $n=0$ term is just $0$ anyway so we can start it at $n=0$

plugging in:

$$ \sum_{n=0}^\infty n\cdot c_nx^{n} - 3\sum_{n=0}^\infty (n+1)\cdot c_{n+1}x^{n} + 2 \sum_{n=0}^\infty c_nx^{n} = 0$$

$$\sum_{n=0}^\infty (nC_n - 3(n+1)C_{n+1} + 2C_n)x^n = 0$$

so for this equation to be true, the terms have to match. Since there is no $x^n$ term on the right, we can set the coefficient term to $0$.

$$(nC_n - 3(n+1)C_{n+1} + 2C_n) = 0$$

$$(C_n(n+2) - 3(n+1)C_{n+1}) = 0$$

$$\frac{(C_n(n+2)}{3(n+1)} = C_{n+1}) $$

Am I on track so far?

From here, I can find the general term right by inspecting specific terms:

$c_0 = c_0$ and $c_1 = \frac{2c_0}{3}$ and $c_2 = \frac{3c_1}{6} = \frac{3 \cdot 2 \cdot c_0}{6 \cdot 3}$ and $c_3 = \frac{4c_2}{9} = \frac{4! \cdot c_0}{9 \cdot 6 \cdot 3}$ and $c_n = \frac{(n+1)! c_0}{3 \cdot 6 \cdot 9 \cdot ... 3n}$

so $$y = \sum_{n=0}^\infty \frac{(n+1)! c_0}{3 \cdot 6 \cdot 9 \cdot ... 3n} x^n$$

Is this right? Is there anything else I can do here?

  • 1
    $\begingroup$ You should get: $$ c_n=\frac{n+1}{3^n}c_0. $$ $\endgroup$ – Wang May 10 at 19:29

You can do this also like this:

You can write it $${y'\over y} = {-2\over x-3}$$

and thus $$(\ln(y))' = (-2\ln(x-3))'\implies \ln y = -2\ln(x-3)+c$$ so $$ y={A\over (x-3)^2}$$

But I'm not sure if it is of any help.

  • 1
    $\begingroup$ I think you have a sign error at the begining. $\endgroup$ – Botond May 10 at 19:28

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