how to Solve homogeneous linear difference equation x(n+2)-2x(n+1)+10x(n)=0,x(0)=1/2,x(1)=3/2

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Introduction : The step by step solution of the question how to Solve homogeneous linear difference equation x(n+2)-2x(n+1)+10x(n)=0,x(0)=1/2,x(1)=3/2 is explained below.

Question : Solve

\[x_{n+2}-2x_{n+1}+10x_{n}=0\]

with initial conditions \(x_0=\frac{1}{2}, x_1 =\frac{3}{2}\)

Solution:

We have given is homogeneous Linear difference equation

\[x_{n+2}-2x_{n+1}+10x_{n}=0\]

To solve a homogeneous linear difference equation we first find it’s characteristic equation

\[r^2-2r+10=0\]

\[r^2-2r+1+9=0\]

\[(r-1)^2-(3i)^2=0\]

\[(r-1-3i)(r-1+3i)=0\]

\[r_1 =1+3i, r_2 =1-3i\]

roots are complex

In complex case General solution of homogeneous Linear difference equation is of the form

\[x_{n}=\left(A\cos\left(n\tan^{-1}\left(\frac{3}{1}\right)\right)+B\sin\left(n\tan^{-1}\left(\frac{3}{1}\right)\right)\right)\]

\[x_{n}=10^{\frac{n}{2}}\left(A\cos\left(n\tan^{-1}\left(\frac{3}{1}\right)\right)+B\sin\left(n\tan^{-1}\left(\frac{3}{1}\right)\right)\right)\]

Now To find constants A and B

Put n=0 and using first initial condition \(x_0 =\frac{1}{2}\)

\[\frac{1}{2}=A\]

and Put n=1 and using second initial condition (x_1 = \frac{3}{2}\)

\[\frac{3}{2}=10^{\frac{1}{2}}\left(A\cos\left(\tan^{-1}\left(\frac{3}{1}\right)\right)+B\sin\left(\tan^{-1}\left(\frac{3}{1}\right)\right)\right)\]

Solving these equations for A, B

\[A = 1, B = \frac{\left(\frac{3}{2\cdot\sqrt{10}}-\cos\left(\tan^{-1}\left(\frac{3}{1}\right)\right)\right)}{\sin\left(\tan^{-1}\left(\frac{3}{1}\right)\right)} = \frac{1}{6}​\]

Therefore specific solution of the given difference equation becomes now

\[x_{n}=10^{\frac{n}{2}}\left(\cos\left(n\tan^{-1}\left(\frac{3}{1}\right)\right)+\frac{1}{6}\sin\left(n\tan^{-1}\left(\frac{3}{1}\right)\right)\right)\]


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