0
$\begingroup$

$\displaystyle y^{'} = {\frac{y}{x}} + 1$
cannot be solved for $y$ as a power series $x$. Solve this equation for $y$ as a power series in powers for $ x-1 $.: Introduce $t=x-1$ as a new independent variable and solve the resulting equation for y as a power series in $t$.)

So far I have proved that it cannot be solved for y as power series in terms of x because you get negative exponents. I did solve for x, which equal t+1.

So, the equation looks like $\displaystyle y^{'} = {\frac{y}{t+1}} + 1$.

Can I multiply both sides by $(t+1)$ so the equation looks like $ \displaystyle (t+1)y^{'} = y + (t+1)$?

My solution so far:

$y^{'} = \displaystyle \sum^{\infty}_{n=1}{n}{c_n}{(t+1)^{n-1}}$
$y = \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}}$

$ \displaystyle (t+1)y^{'} = y + (t+1)$

$(t+1)\displaystyle \sum^{\infty}_{n=1}{n}{c_n}{(t+1)^{n-1}} = \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}} + (t+1)$

$\displaystyle \sum^{\infty}_{n=1}{n}{c_n}{(t+1)^{n}} - \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}} =(t+1)$

$\displaystyle \sum^{\infty}_{n=0}{n}{c_n}{(t+1)^{n}} - \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}} =(t+1)$

$\displaystyle \sum^{\infty}_{n=0}[nc_n-c_n]*(t+1)^n = (t+1)$

$\displaystyle nc_n-c_n = (t+1)$

$\displaystyle c_n(n-1) = (t+1)$

$\displaystyle c_n = \frac {(t+1)}{n-1} $

Substituing x back in we get:

$\displaystyle c_n = \frac {(x)}{n-1} $

Did I do this correctly?

$\endgroup$
0
$\begingroup$

$\displaystyle \sum^{\infty}_{n=0}[nc_n-c_n]*(t+1)^n = (t+1)$

$\displaystyle nc_n-c_n = (t+1)$

$\displaystyle c_n(n-1) = (t+1)$

$\displaystyle c_n = \frac {(t+1)}{n-1} $

Substituing x back in we get:

$\displaystyle c_n = \frac {(x)}{n-1} $

Did I do this correctly?

From this line is is not.

$$ \sum^{\infty}_{n=0}[nc_n-c_n]*(t+1)^n = (t+1) \implies \forall n\neq 1\ \ \ nc_n-c_n = 0 $$

At the end, $c_n$ should not depend on $x$!

Note that you can solve the equation directly with the substitution $$ y(x) = r(x)x $$

$\endgroup$

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.