GATE CSE
First time here? Checkout the FAQ!
x
+3 votes
70 views

find the solution of the recurence relation

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

asked in Mathematical Logic by Active (1.3k points)   | 70 views

2 Answers

+3 votes
Best answer

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

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

answered by Boss (8.1k points)  
selected by
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.

+4 votes

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 $a_{n}=[11/6].3^{n}-n-3/2.$

answered by Veteran (10.7k points)  


Top Users Mar 2017
  1. rude

    5118 Points

  2. sh!va

    3054 Points

  3. Rahul Jain25

    2920 Points

  4. Kapil

    2730 Points

  5. Debashish Deka

    2602 Points

  6. 2018

    1574 Points

  7. Vignesh Sekar

    1422 Points

  8. Akriti sood

    1382 Points

  9. Bikram

    1350 Points

  10. Sanjay Sharma

    1128 Points

Monthly Topper: Rs. 500 gift card

21,531 questions
26,861 answers
61,189 comments
23,206 users