Use series methods to find the solution corresponding to....

[shamieh]

Use series methods to find the solution corresponding to $a_0 = 1$ for the equation: $(x+1)y' - y = 0$

I just need someone to verify if my solution is correct.

I ended up with the recursive formula being: $a_{n+1} = a_n$ and I ended up with $y_1 = 1 + 0 + 0 ...$ and $y_2 = x + x^2 + x^3 + x^4...$ and $y = C_1y_1 + C_2y_2$

View attachment 4318

 

[Prove It]

Do you really have to use series for this? You can get an EXACT solution quite easily by separating the variables or using an integrating factor...

$\displaystyle \begin{align*} \left( x + 1 \right) \, \frac{\mathrm{d}y}{\mathrm{d}x} - y &= 0 \\ \left( x + 1 \right) \, \frac{\mathrm{d}y}{\mathrm{d}x} &= y \\ \frac{1}{y} \, \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{x+1} \\ \int{ \frac{1}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x} \,\mathrm{d}x} &= \int{ \frac{1}{x+1}\,\mathrm{d}x} \\ \int{ \frac{1}{y}\,\mathrm{d}y} &= \ln{|x+1|} + C_1 \\ \ln{|y|} + C_2 &= \ln{|x+1|} + C_1 \\ \ln{|y|} - \ln{|x+1|} &= C_1 - C_2 \\ \ln{ \left| \frac{y}{x+1} \right| } &= C_1 - C_2 \\ \left| \frac{y}{x+1} \right| &= \mathrm{e}^{C_1 - C_2} \\ \frac{y}{x+1} &= C \textrm{ where } C = \pm \mathrm{e}^{C_1 - C_2} \\ y &= C\,\left( x + 1 \right) \end{align*}$

 

[shamieh]

Yes, unfortunately, I have to use the series method as the question specifically states Use series methods to solve.. :(

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Is my solution correct?

 

[shamieh]

So for my final solution I have

recursive formula: $a_{n+1} = \frac{a_n(1-n)}{n+1}$

$y_1 = 1 + 0.. + 0 .. + ...$

so $y = C_1y_1$

 

[shamieh]

UPDATE: SOLVED

The final solution is:

Recursion Formula: $a_{n+1} = \frac{a_n(1-n)}{n+1}$

$\therefore y_1 = 1 + x + 0 + 0 + ... + ...$