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Let's say I have an equation like this:

$Sn = 2 \cdot S(n-1) + 2^{(n+1)} - 2$

Now I know that I need to transform it to look like this:

$S(n+1) - 2 \cdot Sn = 2^{(n+2)} - 2$

From this I can get the characteristic equation: $\lambda - 2$ and from that the characteristic number 2. But how do I find the particular solution? And how do I then add it all together? Any explanations or links to materials are appreciated. I did find some but it did not make sense to me.

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  • $\begingroup$ Please typeset the equations using MathJax. $\endgroup$
    – user65203
    Jan 21, 2019 at 10:12

1 Answer 1

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The homogeneous equation reads

$$S_n=2S_{n-1}$$ and by induction

$$S_n=2^nS_0.$$

Now let us find a particular solution for the term $-2$, and let us try a constant $a$:

$$a=2a-2$$ yields the solution $a=2$.

Now finding a particular solution for the term $2^{n+1}$ is more tricky. You will be tempted to use the Ansatz $b\,2^n$, but as this is of the same form as the homogeneous solution, it won't work:

$$b\,2^n=2b\,2^{n-1}+2^{n+1}$$ has no solution.

Inspired by the theory of linear ODEs, we will try $bn\,2^n$, and

$$bn\,2^n=2b(n-1)2^{n-1}+2^{n+1},$$ giving $b=2$.

The global solution is just the sum of these contributions,

$$S_n=c\,2^n+2n\,2^n+2.$$

Indeed,

$$S_n-2S_{n-1}=c\,2^n-c\,2^n+2n\,2^n-2(n-1)\,2^n+2-4=2^{n+1}-2.$$

You determine $c$ from the initial condition,

$$S_0=c+2.$$

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