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Exercise Set 9E
These questions are considered intermediate to advanced Higher Level.
Question 1
Solve the following difference equation
\begin{align}T_{n+2}-7T_{n+1}+12T_n=6n+13\end{align}
when \(T_1=1\) and \(T_2=5\).
\(T_n=-4(3^n)+2.25(4^n)+n+3\)
Complimentary Solution
\begin{align}x^2-7x+12=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-3)(x-4)=0\end{align}
\(x=3\) or \(x=4\)
\begin{align}\downarrow\end{align}
\begin{align}T_n = p(3^n) + q(4^n)\end{align}
Particular Solution
\(D(n)=6n+13\)
\begin{align}\downarrow\end{align}
\begin{align}T_n = rn+s\end{align}
and
\begin{align}T_{n+1} &= r(n+1)+s\\&=rn+r+s\end{align}
and
\begin{align}T_{n+2} &= r(n+2)+s\\&=rn+2r+s\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(rn+2r+s)-7(rn+r+s)+12(rn+s)=6n+13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}rn+2r+s-7rn-7r-7s+12rn+12s=6n+13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(6r)n+(-5r+6s)=6n+13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6r=6\end{align}
\begin{align}-5r+6s=13\end{align}
\begin{align}\downarrow\end{align}
\(r=1\)
\begin{align}-5(1)+6s=13\end{align}
\begin{align}\downarrow\end{align}
\(r=1\)
\begin{align}6s=18\end{align}
\begin{align}\downarrow\end{align}
\(r=1\)
\begin{align}s=3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_n = n+3\end{align}
Full Solution
\begin{align}T_n=p(3^n)+q(4^n)+n+3\end{align}
\(T_1=1\) and \(T_2=5\)
\begin{align}\downarrow\end{align}
\begin{align}1=p(3^1)+q(4^1)+1+3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3p+4q=-3\end{align}
and
\begin{align}5=p(3^2)+q(4^2)+2+3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}9p+16q=0\end{align}
We therefore have the following two equations with two unknowns
\begin{align}3p+4q=-3\end{align}
\begin{align}9p+16q=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}9p+12q=-9\end{align}
\begin{align}9p+16q=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4q=9\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=2.25\end{align}
and
\begin{align}3p+4q=-3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3p+4(2.25)=-3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3p+9=-3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3p=-12\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p=-4\end{align}
Therefore, the solution to the difference equation is
\begin{align}T_n=-4(3^n)+2.25(4^n)+n+3\end{align}
This is a second order inhomogeneous difference equation and we shall therefore use the corresponding steps to solve it.
Step 1: Find the corresponding complimentary solution with unknown constants.
To do this, we shall use our usual steps to solve the following second order inhomogeneous equation (but without finding the unknown constants)
\begin{align}T_{n+2}-7T_{n+1} + 12T_n = 0\end{align}
Step 1: Solve the corresponding characteristic equation.
The corresponding characteristic equation is given by
\begin{align}x^2-7x+12=0\end{align}
This can be factorised as
\begin{align}(x-3)(x-4)=0\end{align}
and therefore \(x=3\) or \(x=4\).
\[\,\]
Step 2: Write the general term in terms of unknown constants, taking into account whether there is is \(\boldsymbol{1}\) or \(\boldsymbol{2}\) distinct roots.
The characteristic equation has two distinct roots \(\alpha=3\) and \(\beta=4\).
The solution to the homogeneous difference equation therefore has the form
\begin{align}T_n = p(3^n) + q(4^n)\end{align}
The complimentary solution is therefore
\begin{align}T_n = p(3^n) + q(4^n)\end{align}
\[\,\]
Step 2: Write the form of the particular solution based on the form of \(\boldsymbol{D(n)}\).
As \(D(n)=6n+13\), the particular solution is of the form
\begin{align}T_n = rn+s\end{align}
\[\,\]
Step 3: Find the unknown constants within the particular solution by using the difference equation.
Since \(T_n= rn+s\), we also know that
\begin{align}T_{n+1} &= r(n+1)+s\\&=rn+r+s\end{align}
and
\begin{align}T_{n+2} &= r(n+2)+s\\&=rn+2r+s\end{align}
Inserting these three terms into the difference equation, we obtain
\begin{align}(rn+2r+s)-7(rn+r+s)+12(rn+s)=6n+13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}rn+2r+s-7rn-7r-7s+12rn+12s=6n+13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(6r)n+(-5r+6s)=6n+13\end{align}
By comparing the left hand side to the right hand side, we obtain the following equations
\begin{align}6r=6\end{align}
\begin{align}-5r+6s=13\end{align}
The solution to the first equation is \(r=1\) which, upon inserting into the second equation, gives
\begin{align}-5(1)+6s=13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6s=18\end{align}
\begin{align}\downarrow\end{align}
\begin{align}s=3\end{align}
The particular solution is therefore
\begin{align}T_n = n+3\end{align}
\[\,\]
Step 4: Write the full solution.
The full solution therefore has the form
\begin{align}T_n=p(3^n)+q(4^n)+n+3\end{align}
\[\,\]
Step 5: Find the unknown constants in the full solution using the given terms.
To find both \(p\) and \(q\), we will use the fact that we know that \(T_1=1\) and \(T_2=5\).
Inserting that \(T_1=1\) gives
\begin{align}1=p(3^1)+q(4^1)+1+3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3p+4q=-3\end{align}
Inserting that \(T_2=5\) instead gives
\begin{align}5=p(3^2)+q(4^2)+2+3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}9p+16q=0\end{align}
We therefore have the following two equations with two unknowns
\begin{align}3p+4q=-3\end{align}
\begin{align}9p+16q=0\end{align}
Multiplying the first equation by \(3\)
\begin{align}9p+12q=-9\end{align}
\begin{align}9p+16q=0\end{align}
and then subtracting it from the second equation gives
\begin{align}4q=9\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=2.25\end{align}
Inserting this into e.g. the following
\begin{align}3p+4q=-3\end{align}
then gives
\begin{align}3p+4(2.25)=-3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3p+9=-3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3p=-12\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p=-4\end{align}
Therefore, the solution to the difference equation is
\begin{align}T_n=-4(3^n)+2.25(4^n)+n+3\end{align}
Question 2
Solve the following difference equation
\begin{align}T_{n+1}-4T_{n}-5T_{n-1}=16n^2+8\end{align}
when \(T_0=6\) and \(T_1=11\).
\(T_n = \dfrac{23}{6}(-1)^n + \dfrac{53}{12}(5^n)-2n^2-3n-2.25\)
Complimentary Solution
\begin{align}x^2-4x-5=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x+1)(x-5)=0\end{align}
\(x=-1\) or \(x=5\)
\begin{align}\downarrow\end{align}
\begin{align}T_n &= p(-1)^n + q(5^n)\end{align}
Particular Solution
\(D(n)=16n^2+18\)
\begin{align}T_n = rn^2+sn+t\end{align}
and
\begin{align}T_{n+1} &= r(n+1)^2+s(n+1)+t\\&=rn^2+(2r+s)n+r+s+t\end{align}
and
\begin{align}T_{n-1} &= r(n-1)^2+s(n-1)+t\\&=rn^2-(2r-s)n+r-s+t\end{align}
\begin{align}\downarrow\end{align}
\begin{align}[rn^2+(2r+s)n+r+s+t]-4[rn^2+sn+t]-5[rn^2-(2r-s)n+r-s+t]=16n^2+8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}rn^2+(2r+s)n+r+s+t-4rn^2-4sn-4t-5rn^2+(10r-5s)n-5r+5s-5t=16n^2+8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(-8r)n^2+(12r-8s)n-4r+6s-8t=16n^2+8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-8r=16\end{align}
\begin{align}12r-8s=0\end{align}
\begin{align}-4r+6s-8t=8\end{align}
The solution to the first equation is \(r=-2\) which, upon inserting into the second equation, gives
\begin{align}r=-2\end{align}
\begin{align}12r-8s=0\end{align}
\begin{align}-4r+6s-8t=8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=-2\end{align}
\begin{align}8s=-24\end{align}
\begin{align}6s-8t=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=-2\end{align}
\begin{align}s=-3\end{align}
\begin{align}6(-3)-8t=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=-2\end{align}
\begin{align}s=-3\end{align}
\begin{align}-8t=18\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=-2\end{align}
\begin{align}s=-3\end{align}
\begin{align}t=-2.25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_n = -2n^2-3n-2.25\end{align}
Full Solution
\begin{align}T_n = p(-1)^n + q(5^n)-2n^2-3n-2.25\end{align}
\(T_0=6\) and \(T_1=11\)
\begin{align}\downarrow\end{align}
\begin{align}6=p(-1)^0 + q(5^0)-2(0^2)-3(0)-2.25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p+q=8.25\end{align}
and
\begin{align}11= p(-1)^1 + q(5^1)-2(1^2)-3(1)-2.25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-p+5q=18.25\end{align}
We therefore have the following two equations with two unknowns
\begin{align}p+q=8.25\end{align}
\begin{align}-p+5q=18.25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6q=26.5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=\frac{53}{12}\end{align}
and
\begin{align}p+q=8.25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p&=8.25-q\\&=8.25-\frac{53}{12}\\&=\frac{23}{6}\end{align}
Therefore, the solution to the difference equation is
\begin{align}T_n = \frac{23}{6}(-1)^n + \frac{53}{12}(5^n)-2n^2-3n-2.25\end{align}
This is a second order inhomogeneous difference equation and we shall therefore use the corresponding steps to solve it.
Step 1: Find the corresponding complimentary solution with unknown constants.
To do this, we shall use our usual steps to solve the following second order inhomogeneous equation (but without finding the unknown constants)
\begin{align}T_{n+1}-7T_{n} + 12T_{n-1} = 0\end{align}
Step 1: Solve the corresponding characteristic equation.
The corresponding characteristic equation is given by
\begin{align}x^2-4x-5=0\end{align}
This can be factorised as
\begin{align}(x+1)(x-5)=0\end{align}
and therefore \(x=-1\) or \(x=5\).
\[\,\]
Step 2: Write the general term in terms of unknown constants, taking into account whether there is is \(\boldsymbol{1}\) or \(\boldsymbol{2}\) distinct roots.
The characteristic equation has two distinct roots \(\alpha=-1\) and \(\beta=5\).
The solution to the homogeneous difference equation therefore has the form
\begin{align}T_n &= p(-1)^n + q(5^n)\end{align}
The complimentary solution is therefore
\begin{align}T_n &= p(-1)^n + q(5^n)\end{align}
\[\,\]
Step 2: Write the form of the particular solution based on the form of \(\boldsymbol{D(n)}\).
As \(D(n)=16n^2+18\), the particular solution is of the form
\begin{align}T_n = rn^2+sn+t\end{align}
\[\,\]
Step 3: Find the unknown constants within the particular solution by using the difference equation.
Since \(T_n= rn^2+sn+t\), we also know that
\begin{align}T_{n+1} &= r(n+1)^2+s(n+1)+t\\&=rn^2+(2r+s)n+r^2+s+t\end{align}
and
\begin{align}T_{n-1} &= r(n-1)^2+s(n-1)+t\\&=rn^2-(2r-s)n+r^2-s+t\end{align}
Inserting these three terms into the difference equation, we obtain
\begin{align}[rn^2+(2r+s)n+r^2+s+t]-4[rn^2+sn+t]-5[rn^2-(2r-s)n+r^2-s+t]=16n^2+8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}rn^2+(2r+s)n+r^2+s+t-4rn^2-4sn-4t-5rn^2+(10r-5s)n-5r^2+5s-5t=16n^2+8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(-8r)n^2+(12r-8s)n-4r^2+6s-8t=16n^2+18\end{align}
By comparing the left hand side to the right hand side, we obtain the following equations
\begin{align}-8r=16\end{align}
\begin{align}12r-8s=0\end{align}
\begin{align}-4r^2+6s-8t=18\end{align}
The solution to the first equation is \(r=-2\) which, upon inserting into the second equation, gives
\begin{align}12r-8s=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}8s=-24\end{align}
\begin{align}\downarrow\end{align}
\begin{align}s=-3\end{align}
Finally, inserting both of these into the third equation gives
\begin{align}-4(-2)^2+6(-3)-8t=18\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-16-18-8t=18\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-8t=52\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=-6.5\end{align}
The particular solution is therefore
\begin{align}T_n = -2n^2-3n-6.5\end{align}
\[\,\]
Step 4: Write the full solution.
The full solution therefore has the form
\begin{align}T_n = p(-1)^n + q(5^n)-2n^2-3n-6.5\end{align}
\[\,\]
Step 5: Find the unknown constants in the full solution using the given terms.
To find both \(p\) and \(q\), we will use the fact that we know that \(T_0=6\) and \(T_1=11\).
Inserting that \(T_0=6\) gives
\begin{align}6=p(-1)^0 + q(5^0)-2(0^2)-3(0)-6.5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p+q=12.5\end{align}
Inserting that \(T_1=11\) instead gives
\begin{align}11= p(-1)^1 + q(5^1)-2(1^2)-3(1)-6.5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-p+5q=22.5\end{align}
We therefore have the following two equations with two unknowns
\begin{align}p+q=12.5\end{align}
\begin{align}-p+5q=22.5\end{align}
Subtracting the first equation from the second equation, we obtain
\begin{align}4q=10\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=2.5\end{align}
Inserting this into e.g. the following
\begin{align}p+q=12.5\end{align}
then gives
\begin{align}p&=12.5-q\\&=12.5-2.5\\&=10\end{align}
Therefore, the solution to the difference equation is
\begin{align}T_n = 10(-1)^n + 2.5(5^n)-2n^2-3n-6.5\end{align}
Question 3
Solve the following difference equation
\begin{align}T_{n+2}-11T_{n+1}+10T_n=2^n\end{align}
when \(T_1=1\) and \(T_2=2\).
\(T_n=\dfrac{248}{297}+\dfrac{31}{2970}(10^n)+\dfrac{2^n}{33}\)
Complimentary Solution
\begin{align}x^2-11x+10=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-1)(x-10)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=1\) or \(x=10\)
\begin{align}\downarrow\end{align}
\begin{align}T_n &= p(1^n) + q(10^n)\\&=p+q(10^n)\end{align}
Particular Solution
\(D(n)=2^n\)
\begin{align}\downarrow\end{align}
\begin{align}T_n = r(2^n)\end{align}
and
\begin{align}T_{n+1} &= r(2^{n+1})\\&=2r(2^n)\end{align}
and
\begin{align}T_{n+2} &= r(2^{n+2})\\&=4r(2^n)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4r(2^n)-11r(2^n)+40r(2^n)=2^n\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4r-11r+40r=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}33r=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=\frac{1}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_n = \frac{2^n}{33}\end{align}
Full Solution
\begin{align}T_n=p+q(10^n)+\frac{2^n}{33}\end{align}
\(T_1=1\) and \(T_2=2\)
\begin{align}\downarrow\end{align}
\begin{align}1=p+q(10^1)+\frac{2^1}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p+10q=\frac{31}{33}\end{align}
and
\begin{align}2=p+q(10^2)+\frac{2^2}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p+100q=\frac{62}{33}\end{align}
We therefore have the following two equations with two unknowns
\begin{align}p+10q=\frac{31}{33}\end{align}
\begin{align}p+100q=\frac{62}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}90q=\frac{31}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=\frac{31}{2970}\end{align}
and
\begin{align}p+10q=\frac{31}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p+10\left(\frac{31}{2970}\right)=\frac{31}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p&=\frac{31}{33}-10\left(\frac{31}{2970}\right)\\&=\frac{248}{297}\end{align}
Therefore, the solution to the difference equation is
\begin{align}T_n=\frac{248}{297}+\frac{31}{2970}(10^n)+\frac{2^n}{33}\end{align}
This is a second order inhomogeneous difference equation and we shall therefore use the corresponding steps to solve it.
Step 1: Find the corresponding complimentary solution with unknown constants.
To do this, we shall use our usual steps to solve the following second order inhomogeneous equation (but without finding the unknown constants)
\begin{align}T_{n+2}-11T_{n+1} + 10T_n = 0\end{align}
Step 1: Solve the corresponding characteristic equation.
The corresponding characteristic equation is given by
\begin{align}x^2-11x+10=0\end{align}
This can be factorised as
\begin{align}(x-1)(x-10)=0\end{align}
and therefore \(x=1\) or \(x=10\).
\[\,\]
Step 2: Write the general term in terms of unknown constants, taking into account whether there is is \(\boldsymbol{1}\) or \(\boldsymbol{2}\) distinct roots.
The characteristic equation has two distinct roots \(\alpha=1\) and \(\beta=10\).
The solution to the homogeneous difference equation therefore has the form
\begin{align}T_n &= p(1^n) + q(10^n)\\&=p+q(10^n)\end{align}
(as \(1^n=1\) for all \(n\).)
The complimentary solution is therefore
\begin{align}T_n=p+q(10^n)\end{align}
\[\,\]
Step 2: Write the form of the particular solution based on the form of \(\boldsymbol{D(n)}\).
As \(D(n)=2^n\), the particular solution is of the form
\begin{align}T_n = r(2^n)\end{align}
\[\,\]
Step 3: Find the unknown constants within the particular solution by using the difference equation.
Since \(T_n= r(2^n)\), we also know that
\begin{align}T_{n+1} &= r(2^{n+1})\\&=2r(2^n)\end{align}
and
\begin{align}T_{n+2} &= r(2^{n+2})\\&=4r(2^n)\end{align}
Inserting these three terms into the difference equation, we obtain
\begin{align}4r(2^n)-11r(2^n)+40r(2^n)=2^n\end{align}
Cancelling the \(2^n\) terms, we obtain:
\begin{align}4r-11r+40r=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}33r=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=\frac{1}{33}\end{align}
The particular solution is therefore
\begin{align}T_n = \frac{2^n}{33}\end{align}
\[\,\]
Step 4: Write the full solution.
The full solution therefore has the form
\begin{align}T_n=p+q(10^n)+\frac{2^n}{33}\end{align}
\[\,\]
Step 5: Find the unknown constants in the full solution using the given terms.
To find both \(p\) and \(q\), we will use the fact that we know that \(T_1=1\) and \(T_2=2\).
Inserting that \(T_1=1\) gives
\begin{align}1=p+q(10^1)+\frac{2^1}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p+10q=\frac{31}{33}\end{align}
Inserting that \(T_2=2\) instead gives
\begin{align}2=p+q(10^2)+\frac{2^2}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p+100q=\frac{62}{33}\end{align}
We therefore have the following two equations with two unknowns
\begin{align}p+10q=\frac{31}{33}\end{align}
\begin{align}p+100q=\frac{62}{33}\end{align}
Subtracting the first equation from the second equation gives
\begin{align}90q=\frac{31}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=\frac{31}{2970}\end{align}
Inserting this into e.g. the following
\begin{align}p+10q=\frac{31}{33}\end{align}
then gives
\begin{align}p+10\left(\frac{31}{2970}\right)=\frac{31}{33}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p&=\frac{31}{33}-10\left(\frac{31}{2970}\right)\\&=\frac{248}{297}\end{align}
Therefore, the solution to the difference equation is
\begin{align}T_n=\frac{248}{297}+\frac{31}{2970}(10^n)+\frac{2^n}{33}\end{align}
Question 4
Solve the following difference equation
\begin{align}T_{n+1}-8T_{n}+12T_{n-1}=4(3^n)\end{align}
when \(T_0=15\) and \(T_1=5\).
\(T_n=\dfrac{97}{4}(2^n)-\dfrac{21}{4}(6^n)-4(3^n)\)
Complimentary Solution
\begin{align}x^2-8x+12=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-2)(x-6)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=2\) or \(x=6\)
\begin{align}\downarrow\end{align}
\begin{align}T_n = p(2^n) + q(6^n)\end{align}
Particular Solution
\(D(n)=4(3^n)\)
\begin{align}\downarrow\end{align}
\begin{align}T_n = r(3^n)\end{align}
and
\begin{align}T_{n+1} &= r(3^{n+1})\\&=3r(3^n)\end{align}
and
\begin{align}T_{n-1} &= r(3^{n-1})\\&=\frac{r}{3}(3^n)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3r(3^n)-8r(3^n)+4r(3^n)=4(3^n)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3r-8r+4r=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-r=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=-4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_n = -4(3^n)\end{align}
Full Solution
\begin{align}T_n = p(2^n) + q(6^n)-4(3^n)\end{align}
\(T_0=15\) and \(T_1=5\)
\begin{align}\downarrow\end{align}
\begin{align}15=p(2^0) + q(6^0)-4(3^0)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p+q=19\end{align}
and
\begin{align}5=p(2^1) + q(6^1)-4(3^1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2p+6q=17\end{align}
We therefore have the following two equations with two unknowns
\begin{align}p+q=19\end{align}
\begin{align}2p+6q=17\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2p+2q=38\end{align}
\begin{align}2p+6q=17\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4q=-21\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=-\frac{21}{4}\end{align}
and
\begin{align}p+q=19\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p&=19-q\\&=19-\left(-\frac{21}{4}\right)\\&=\frac{97}{4}\end{align}
Therefore, the solution to the difference equation is
\begin{align}T_n = \frac{97}{4}(2^n)-\frac{21}{4}(6^n)-4(3^n)\end{align}
This is a second order inhomogeneous difference equation and we shall therefore use the corresponding steps to solve it.
Step 1: Find the corresponding complimentary solution with unknown constants.
To do this, we shall use our usual steps to solve the following second order inhomogeneous equation (but without finding the unknown constants)
\begin{align}T_{n+1}-8T_{n} + 12T_{n-1} = 0\end{align}
Step 1: Solve the corresponding characteristic equation.
The corresponding characteristic equation is given by
\begin{align}x^2-8x+12=0\end{align}
This can be factorised as
\begin{align}(x-2)(x-6)=0\end{align}
and therefore \(x=2\) or \(x=6\).
\[\,\]
Step 2: Write the general term in terms of unknown constants, taking into account whether there is is \(\boldsymbol{1}\) or \(\boldsymbol{2}\) distinct roots.
The characteristic equation has two distinct roots \(\alpha=2\) and \(\beta=6\).
The solution to the homogeneous difference equation therefore has the form
\begin{align}T_n = p(2^n) + q(6^n)\end{align}
The complimentary solution is therefore
\begin{align}T_n = p(2^n) + q(6^n)\end{align}
\[\,\]
Step 2: Write the form of the particular solution based on the form of \(\boldsymbol{D(n)}\).
As \(D(n)=4(3^n)\), the particular solution is of the form
\begin{align}T_n = r(3^n)\end{align}
\[\,\]
Step 3: Find the unknown constants within the particular solution by using the difference equation.
Since \(T_n= r(3^n)\), we also know that
\begin{align}T_{n+1} &= r(3^{n+1})\\&=3r(3^n)\end{align}
and
\begin{align}T_{n-1} &= r(3^{n-1})\\&=\frac{r}{3}(3^n)\end{align}
Inserting these three terms into the difference equation, we obtain
\begin{align}3r(3^n)-8r(3^n)+4r(3^n)=4(3^n)\end{align}
Cancelling the \(3^n\) terms, we obtain:
\begin{align}3r-8r+4r=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-r=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=-4\end{align}
The particular solution is therefore
\begin{align}T_n = -4(3^n)\end{align}
\[\,\]
Step 4: Write the full solution.
The full solution therefore has the form
\begin{align}T_n = p(2^n) + q(6^n)-4(3^n)\end{align}
\[\,\]
Step 5: Find the unknown constants in the full solution using the given terms.
To find both \(p\) and \(q\), we will use the fact that we know that \(T_0=15\) and \(T_1=5\).
Inserting that \(T_0=15\) gives
\begin{align}15=p(2^0) + q(6^0)-4(3^0)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p+q=19\end{align}
Inserting that \(T_1=5\) instead gives
\begin{align}5=p(2^1) + q(6^1)-4(3^1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2p+6q=17\end{align}
We therefore have the following two equations with two unknowns
\begin{align}p+q=19\end{align}
\begin{align}2p+6q=17\end{align}
Multiplying the first equation by \(2\)
\begin{align}2p+2q=38\end{align}
\begin{align}2p+6q=17\end{align}
and subtracting the first equation from the second equation gives
\begin{align}4q=-21\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=-\frac{21}{4}\end{align}
Inserting this into e.g. the following
\begin{align}p+q=19\end{align}
then gives
\begin{align}p&=19-q\\&=19-\left(-\frac{21}{4}\right)\\&=\frac{97}{4}\end{align}
Therefore, the solution to the difference equation is
\begin{align}T_n = \frac{97}{4}(2^n)-\frac{21}{4}(6^n)-4(3^n)\end{align}
Question 5
Solve the following difference equation
\begin{align}T_{n+2}-3T_{n+1}-28T_n=10\end{align}
when \(T_1=3\) and \(T_2=4\).
\(T_n=-\dfrac{19}{44}(-4)^n+\dfrac{53}{231}(7^n)-\dfrac{1}{3}\)
Complimentary Solution
\begin{align}x^2-3x-28=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x+4)(x-7)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=-4\) or \(x=7\)
\begin{align}T_n = p(-4)^n + q(7^n)\end{align}
Particular Solution
\(D(n)=10\)
\begin{align}\downarrow\end{align}
\(T_n=r\), \(T_{n+1}=r\) and \(T_{n+2}=r\)
\begin{align}\downarrow\end{align}
\begin{align}r-3r-28r=10\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-30r=10\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=-\frac{1}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_n = -\frac{1}{3}\end{align}
Full Solution
\begin{align}T_n = p(-4)^n + q(7^n)-\frac{1}{3}\end{align}
\(T_1=3\) and \(T_2=4\)
\begin{align}\downarrow\end{align}
\begin{align}3=p(-4)^1 + q(7^1)-\frac{1}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-4p+7q=\frac{10}{3}\end{align}
and
\begin{align}4=p(-4)^2 + q(7^2)-\frac{1}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}16p+49q=\frac{13}{3}\end{align}
We therefore have the following two equations with two unknowns
\begin{align}-4p+7q=\frac{10}{3}\end{align}
\begin{align}16p+49q=\frac{13}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-16p+28q=\frac{40}{3}\end{align}
\begin{align}16p+49q=\frac{13}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}77q=\frac{53}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=\frac{53}{231}\end{align}
and
\begin{align}-4p+7q=\frac{10}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-4p+7\left(\frac{53}{231}\right)=\frac{10}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-4p=\frac{19}{11}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p=-\frac{19}{44}\end{align}
Therefore, the solution to the difference equation is
\begin{align}T_n=-\frac{19}{44}(-4)^n+\frac{53}{231}(7^n)-\frac{1}{3}\end{align}
This is a second order inhomogeneous difference equation and we shall therefore use the corresponding steps to solve it.
Step 1: Find the corresponding complimentary solution with unknown constants.
To do this, we shall use our usual steps to solve the following second order inhomogeneous equation (but without finding the unknown constants)
\begin{align}T_{n+2}-3T_{n+1}-28T_n=0\end{align}
Step 1: Solve the corresponding characteristic equation.
The corresponding characteristic equation is given by
\begin{align}x^2-3x-28=0\end{align}
This can be factorised as
\begin{align}(x+4)(x-7)=0\end{align}
and therefore \(x=-4\) or \(x=7\).
\[\,\]
Step 2: Write the general term in terms of unknown constants, taking into account whether there is is \(\boldsymbol{1}\) or \(\boldsymbol{2}\) distinct roots.
The characteristic equation has two distinct roots \(\alpha=-4\) and \(\beta=7\).
The solution to the homogeneous difference equation therefore has the form
\begin{align}T_n = p(-4)^n + q(7^n)\end{align}
The complimentary solution is therefore
\begin{align}T_n = p(-4)^n + q(7^n)\end{align}
\[\,\]
Step 2: Write the form of the particular solution based on the form of \(\boldsymbol{D(n)}\).
As \(D(n)=10\), the particular solution is of the form
\begin{align}T_n=r\end{align}
\[\,\]
Step 3: Find the unknown constants within the particular solution by using the difference equation.
Since \(T_n=r\), we also know that \(T_{n+1}=r\) and \(T_{n+2}=r\).
Inserting these three terms into the difference equation, we obtain
\begin{align}r-3r-28r=10\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-30r=10\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=-\frac{1}{3}\end{align}
The particular solution is therefore
\begin{align}T_n = -\frac{1}{3}\end{align}
\[\,\]
Step 4: Write the full solution.
The full solution therefore has the form
\begin{align}T_n = p(-4)^n + q(7^n)-\frac{1}{3}\end{align}
\[\,\]
Step 5: Find the unknown constants in the full solution using the given terms.
To find both \(p\) and \(q\), we will use the fact that we know that \(T_1=3\) and \(T_2=4\).
Inserting that \(T_1=3\) gives
\begin{align}3=p(-4)^1 + q(7^1)-\frac{1}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-4p+7q=\frac{10}{3}\end{align}
Inserting that \(T_2=4\) instead gives
\begin{align}4=p(-4)^2 + q(7^2)-\frac{1}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}16p+49q=\frac{13}{3}\end{align}
We therefore have the following two equations with two unknowns
\begin{align}-4p+7q=\frac{10}{3}\end{align}
\begin{align}16p+49q=\frac{13}{3}\end{align}
Multiplying the first equation by \(4\)
\begin{align}-16p+28q=\frac{40}{3}\end{align}
\begin{align}16p+49q=\frac{13}{3}\end{align}
and then adding both equations together gives
\begin{align}77q=\frac{53}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=\frac{53}{231}\end{align}
Inserting this into e.g. the following
\begin{align}-4p+7q=\frac{10}{3}\end{align}
then gives
\begin{align}-4p+7\left(\frac{53}{231}\right)=\frac{10}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-4p=\frac{19}{11}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p=-\frac{19}{44}\end{align}
Therefore, the solution to the difference equation is
\begin{align}T_n=-\frac{19}{44}(-4)^n+\frac{53}{231}(7^n)-\frac{1}{3}\end{align}
