# Thread: Related to power series (Differiential Equations)

1. ## Power series homogenous eq (Differiential Equations)

Non-homogeneous differential equation

(x2 + 1)y'' + 3xy' - 4y = cosh(x)

A. Find two linearly independent power series solutions to the homogeneous equation
(x2 + 1)y'' + 3xy' - 4y = 0

B. Find a particular solution to the original differential equation by using the MacLaurin series for f(x) = cosh(x), and give the general solution y = yh + yp

NOTE: Use the recurrence relation to find the first 4 non-zero terms of the series.

I am just confused at where to start. Please tell me how I should start or give hints. After I've read all the advice I'll come back with my solution and look for more help because this section confuses the heck out of me.

Thanks.

2. Originally Posted by Kyeong
Non-homogeneous differential equation

(x2 + 1)y'' + 3xy' - 4y = cosh(x)

A. Find two linearly independent power series solutions to the homogeneous equation
(x2 + 1)y'' + 3xy' - 4y = 0
as far as methodology goes, see here

there are also three links given in post #2, it may do you well to follow them. see if you get the idea of how to go about solving these.

B. Find a particular solution to the original differential equation by using the MacLaurin series for f(x) = cosh(x), and give the general solution y = yh + yp

NOTE: Use the recurrence relation to find the first 4 non-zero terms of the series.
Let $y_p = A \cosh x + B \sinh x$ (of course you would write these as power series).

find $y_p'$ and $y_p''$ and plug them into the original diff eq. on the right, write the power series for $\cosh x$. then, continue much as you did in solving the homogeneous case, but this time, you will equate the coefficients of like powers on both sides, as opposed to making the coefficients zero.

good luck, come back if you have problems

3. Originally Posted by Kyeong
A. Find two linearly independent power series solutions to the homogeneous equation
(x2 + 1)y'' + 3xy' - 4y = 0
Let us start with this. We know there exists a solution to this equation $\Sigma_{n=0}^{\infty} a_nx^n$ (with infinite radius of convergence).
Substituting we get,
$(x^2+1)\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} + 3x\sum_{n=1}^{\infty} na_nx^{n-1}-4\sum_{n=0}^{\infty}a_nx^n=0$.
Multiply out,
$\sum_{n=2}^{\infty}n(n-1)a_nx^n + \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} + 3\sum_{n=1}^{\infty} na_nx^n -4\sum_{n=0}^{\infty}a_nx^n=0$.
Change the index on the 2nd $\Sigma$,
$\sum_{n=2}^{\infty}n(n-1)a_nx^n + \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n + 3\sum_{n=1}^{\infty} na_nx^n -4\sum_{n=0}^{\infty}a_nx^n=0$.
Evaluting the 2nd sum at $n=0,1$ the 3rd at $n=1$ the 4th at $n=0,1$ and combining,
$(2a_2-4a_0)+(6a_1+3a_1-4a_1)x+\sum_{n=2}^{\infty} [n(n-1)a_n+(n+2)(n+1)a_{n+2} + 3na_n - 4a_n]x^n = 0$.
But this can be written as,
$\sum_{n=0}^{\infty} [n(n-1)a_n+(n+2)(n+1)a_{n+2} + 3na_n - 4a_n]x^n = 0$.
This simplifies to,
$\sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} + (n^2+2n-4)a_n]x^n = 0$.
Therefore the recurrence relation is,
$\boxed{a_{n+2} = - \tfrac{n^2+2n-4}{n^2+3n+2}a_n} ~ \ ~ n\geq 0$.