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find the solution of the recurence relation

a=3an-1 + 2n  initial conditon is given as a1=3 ?

Your question is of the form non linear homogeneous solution with constant coefficient.

so

$a_{n}=a^{\left(h\right)}_{n}+a^{\left(p\right)}_{n}$

where ,$\text{h=Homogeneous and p= polynomial}$

$\text{solve } \rightarrow a^{\left(h\right)}_{n}$

characteristics equation will be

$r=3 \Rightarrow root=3$

so,$a^{\left(h\right)}_{n}=\alpha *3^{n}$

now solve $P_{n}=a^{\left(p\right)}_{n}$

Let $Q_{n} \: \text{be our trial equation },Q_{n}=cn+d$

Then our equation

$a_{n}=3*a_{n-1}+2 \: \text{becomes} \left ( c*n+d \right )=3*\left ( c*\left ( n-1 \right )*d \right )$

$\Rightarrow 2cn+2n+2d-3c=0$

$\Rightarrow n \left (2c+2\right )+2d-3c=0$

solving/comparing both sides,

$c=-1,d= - \frac {3}{2}$

Now assemble all things you obtained.

$a_{n}=a^{\left(h\right)}_{n}+a^{\left(p\right)}_{n}$

$a_{n}=\alpha *3^{n}+\left ( cn+d \right )$

$a_{n}=\alpha *3^{n}-n-\frac{3}{2}$

put value of $a_{1}=3,3=\alpha *3^{1}-1-\frac{3}{2},\alpha =\frac{11}{6}$

$a_{n}=\frac{11}{6} *3^{n}-n-\frac{3}{2} \text{is your answer}$

selected
why u take particular solution as : cn+d

for poylnomials of first degree take

n=Cn+d

n^2=Cn^2+bn+a.

2^n=A.2^n.

It is non homogeneous solution.so first find homogenous and find particular solution..

$a_{n}-3a_{n-1}$=2n.---------------------(i)

Case I) finding homogeneous part.

i.e)$a_{n}-3a_{n-1}$=0.

t-3=0

t=3.(it is root)

Solution is $a_{n}$=p.$3^{n}$.

Case II) finding non homogeneous part.

Non homogenneous part is polynomial so $a_{n}$=An+b.$-----------(ii) sub this equation in (i) Then(An+c)-3(A(n-1)+c)=2n n(-2A)+3A-2c=2n. compare both sides then A=-1 and c=-3/2. particular solution is=Homogeneous +non homogeneous .$a_{n}=p.3^{n}-n-3/2.$. Given that a1=3. then 3=3p-1-3/2. p=11/6. substitute in above equation$a_{n}=[11/6].3^{n}-n-3/2.\$