# solving differential equation $(x-3)y' + 2y = 0$

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?

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

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 $$y={A\over (x-3)^2}$$