Solve the second-order difference equation

y[k+2]-5 y[k+1]+6 y[k]=0 y[0]=0 y[1]=2

Question

Solve the second-order difference equation

y[k+2]-5 y[k+1]+6 y[k]=0 y[0]=0 y[1]=2

Accepted Answer

Taking the z transform of both sides of the equation and using the properties of linearity we have

[latex]\mathcal{Z}\{y[k+2]\}-5 \mathcal{Z}\{y[k+1]\}+6 \mathcal{Z}\{y[k]\}=0[/latex]

From the first shift theorem we have

[latex]z^2 Y(z)-z^2 y[0]-z y[1]-5(z Y(z)-z y[0])+6 Y(z)=0[/latex]

where

[latex]Y(z)=\mathcal{Z}\{y[k]\}[/latex]

Substituting values for the conditions gives

[latex]\begin{aligned} z^2 Y(z)-0 z^2-2 z-5(z Y(z)-0 z)+6 Y(z) & =0 \ z^2 Y(z)-5 z Y(z)+6 Y(z) & =2 z \end{aligned}[/latex]

Hence

[latex]\left(z^2-5 z+6 ight) Y(z)=2 z[/latex]

so that

[latex]Y(z)=\mathcal{Z}\{y[k]\}=\frac{2 z}{(z-2)(z-3)}[/latex]

Dividing both sides by z gives

[latex]\frac{Y(z)}{z}=\frac{2}{(z-2)(z-3)}[/latex]

Expressing the r.h.s. in partial fractions yields

[latex]\begin{aligned} & \frac{Y(z)}{z}=\frac{2}{z-3}-\frac{2}{z-2} \ & Y(z)=\frac{2 z}{z-3}-\frac{2 z}{z-2} \end{aligned}[/latex]

Inverting gives the solution to the difference equation:

[latex]y[k]=2\left(3^k ight)-2\left(2^k ight)[/latex]

Question 22.36: Solve the second-order difference equation y[k+2]-5 y[k+1]+......