0
$\begingroup$

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?

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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.