# particular solution for non-homogeneous equation

• May 18th 2011, 07:18 PM
break
particular solution for non-homogeneous equation
hello, i just started learning about this and i'm a little bit confused on how to find the particular solution for these 2 non-homogeneous equations.

$S_n = S_{(n-1)}+2n$ and $g_{n} = 5g_{(n-1)} - 6g_{(n-2)} + n + 5$

I know how to find a particular solution for an equation like $g_{n} = g_{(n-1)} - 2g_{(n-2)} + 2$ which is -1, but when there is a variable n instead of just a constant at the end, i get confused.

can someone help me get started? thanks!
• May 18th 2011, 07:53 PM
chisigma
Quote:

Originally Posted by break
hello, i just started learning about this and i'm a little bit confused on how to find the particular solution for these 2 non-homogeneous equations.

$S_n = S_{(n-1)}+2n$ and $g_{n} = 5g_{(n-1)} - 6g_{(n-2)} + n + 5$

I know how to find a particular solution for an equation like $g_{n} = g_{(n-1)} - 2g_{(n-2)} + 2$ which is -1, but when there is a variable n instead of just a constant at the end, i get confused.

can someone help me get started? thanks!

The first recursive relation in 'standard form' is written as...

$s_{n+1}= s_{n} + 2\ (n+1)$ (1)

If the 'initial condition' is specified as $s_{0}=a$ , then the solution is easy to find...

$s_{1}= a + 2$

$s_{1}= a + 2+4$

...

$s_{n}= a + 2+4+...+2^{n}= a + 2\ (2^{n}-1)$ (1)

... and the 'particular solution' You are searching for is...

$\delta_{n}= 2\ (2^{n}-1)$ (2)

Kind regards

$\chi$ $\sigma$