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TBLOG 5:48 AM 0


Sheet-7


Machine generated alternative text:
Consider the boundary value problem 
If and are continuous on 10, Il, then 
b
Machine generated alternative text:
Answer : 
Explanation : 
Auxiliary equation: 
4• Sin 
Using initial conditions 
u COS 
hence (a) is correct 
hence (b) is true




Machine generated alternative text:
Consider the boundary value problem 
X If u are 
continuous on 10,11 then 
b 
(x)dx



Machine generated alternative text:
Answer : (b),(c),(d) 
Explanation : 
Auxiliary equation: 
COS ('LAO Sin 
is arbitrary 
u'2(X) —C; 
Therefore, (b)(c) are true 
R.H.S



Machine generated alternative text:
For the boundary value problem 
. There exists an 
eigenvalue kfor which there 
corresponds an eigenfunction in (O, 1) 
that 
Does not change sign. 
Changes sign. 
Is positive 
Is negative



Machine generated alternative text:
Answer : 
Explanation : 
Regular SLP 
Case (i): Let 
2k+1 
I J ; (kFNu(o)) are eigen values 
And eigen function 
sin k+— 
Also, eigen functions corresponding to 
distinct eigen values are orthogonal 
Since the equation is self adjoint, only 
real eigen values

Machine generated alternative text:
Let be two eigen functions 
corresponding to distinct eigenvalues 
of the Sturm-Liouville problem 
With boundary 
condition where 
positive and on (0,1). Which of the 
following is NOT true? 
b 
q (x) y, (Dar —O 
(x) and„G) are linearly 
independent. 
must vanish at most once on 
must vanish at least once on


Machine generated alternative text:
Answer : 
Explanation : 
This is a self adjoint regular SLP. So 
corresponding to least eigen value, 
eigen function does not change its 
sign.hence (a) is correct. 
Also iff fis an eigen function then -fis 
also an eigen function, hence (c),(d) are 
correct. 
Eigen function corresponding to eigen 
values other than least eigen value 
change sign, hence (b) is correct.




Machine generated alternative text:
Answer : (d) 
Explanation : 
on (0,1) 
has no zero in (0,1) 
it is eigen function corresponding to 
least eigen value 
Y, corresponds to any ether eigen 
value except the least 
Hence Y2 must have atleast one zero is
Machine generated alternative text:
(7) 
Consider the boundary value problem 
(BVP) on 10, with 
boundary conditions Then 
the above BVP has 
b 
Only trivial solution for every 
Has a unique solution for all 
Has non-trivial solutions for a 
countable set of values of 
Has a non-trivial solution for every

Machine generated alternative text:
For the boundary value problem, 
then for some 
eigen value x. there corresponds 
b 
Only one eigen function 
Two eigen functions 
Two linearly independent eigen 
functions 
Two orthogonal eigen functions



Machine generated alternative text:
Answer : (c) 
Explanation : 
on 
_V(O) y (27t) O 
For Self adjoint B.V.P. The set of eigen 
values are countably infinite and 
unbounded above. 
If (a),(b) are true —set of eigen values is 
empty ( hence bounded) 
If (d) is true set of eigen values will 
become uncountable.

Machine generated alternative text:
The boundary value problem 
has non-trivial 
solution for 
b 
All negative value of k 
All the values of


Machine generated alternative text:
Answer : 
Explanation : 
Every non zero constant multiple of 
eigen function is an eigen function 
We have infinite choices for c if field is 
infinite. On a finite field, we have finite 
number of eigen vectors. 
Our field is R or c 
set of eigen functions is infinite 
For periodic SLP 
; for smallest 
; for other



Machine generated alternative text:
Answer : (c) 
Explanation : 
Case (i) : b0,The eigen values are 
Case (ii) eco , The eigen value are 
Case (iii) 
(Non trivial)



Machine generated alternative text:
(11) 
The Sturm-Liouville problem: 
has its eigenvectors 
given by 
cos„x; where 
'I-OX


Machine generated alternative text:
(12) 
The eigen values of the Sturm Liouville 
system are 
b 
"2 
None of these


Machine generated alternative text:
Answer : (d) 
Explanation : 
cos Sin 
and 
2 c, COS x


Machine generated alternative text:
Answer : (b) 
Explanation : 
Regular SLP 
Case (i): Let 
J ; (n-N) are eigen values 
Case (ii) 
is not an eigen value. 
Case (iii) 
Let



Machine generated alternative text:
(13) 
Let n be a nonnegative integer. The 
eigen values of the Sturm-Liouville 
problemQY4kY20, With boundary 
are 
conditions 
nr2 
n2

Machine generated alternative text:
(14) 
Consider the Sturm Liouville 
problems: with Y'(O-O 
and then the set eigen-valuesk, 
satisfy


Machine generated alternative text:
Answer : (d) 
Explanation : 
For 
COS S in (we) 
—cos Sin 
sin 
(1) and (2) 
I —eos(2wt) —sin(2wt) 
Sin I —cos 
exist 
only trivial solution. 
For non trivial solution, 
; 10) 
if 
Using initial conditions 
...(1)



Machine generated alternative text:
Answer : (a) 
Explanation : 
y' +2xy'Ar2y-o 
Self adjoint regular B.V.P. 
If eigen values are bounded above 
If eigen values are bounded above 
If there may be infinite negative 
eigen values 
The set of eigen values is never 
bounded above and the set of negative 
eigen values is FINITE. 
b, c, d are discarded



Machine generated alternative text:
(15) 
The set of all eigen values of the Sturm 
Liouville problem 
is given by



Machine generated alternative text:
Answer : (d) 
Explanation : 
For 
COS Sin 
1—4n2 ; 
trivialsolution) 
is also eigen value. 
Eigen values :

Machine generated alternative text:
(17) 
The difference between the least two 
eigenvalues of the boundary value 
problem 
such that 
is 
Enter Your Response


Machine generated alternative text:
Answer : 
Explanation : 
The eigen values of this system are 
(2n-l)' 
For k,-1/4 
For n-2, 
Hence



Machine generated alternative text:
(18) 
and where and 
are continuous and 
on then 
b 
The given problem has periodic 
boundary conditions 
There exists a sequence of eigen 
values 
s.t. 
The Eigen vectors corresponding to 
some eigen value change its sign in 
The Eigen space corresponding to 
some eigen value is of two 
dimension


Machine generated alternative text:
(19) 
The eigen values of the strum Liouville 
problem ogxgr, are 
"2



Machine generated alternative text:
Answer : 
Explanation : 
This is periodic SLP and a, b, c, d are 
true by direct properties.
Machine generated alternative text:
Answer : (c) 
Explanation : 
Regular SLP 
Case (i): Let 
• („eN) are eigen values


Machine generated alternative text:
(20) 
The boundary value problem 
b 
has 
Unique eigen value 
Two eigen value 
No eigen value 
Infinite number of eigen value


Machine generated alternative text:
(21) 
For the B.V.P.y„+Ry-0 
to an Eigen value k, there corresponds. 
b 
Only one Eigen function. 
Two Eigen functions. 
Two L.I. Eigen functions. 
Two orthogonal Eigen functions.


Machine generated alternative text:
Answer : (d) 
Explanation : 
4 C2 Sin 
cos ga tan 
Which holds for infinite values of


Machine generated alternative text:
Answer : 
Explanation : 
for least 
Otherwise 
And corresponding to each eigen value 
we have infinite eigen vectors 
a, b are incorrect 
c, d are true if is not least.


Machine generated alternative text:
(22) 
The boundary value problem 
such that 
y(O) y U)) (WII) O 
Does not have any negative eigen 
value 
There exists and an eigen 
function that 
There exists and an eigen 
function that and 
There exists and an eigen 
function YO) such that





Machine generated alternative text:
Answer : 
Explanation : 
We can easily check that this BVP has no 
negative eigen value 
Note:z eigen value and corresponding 
eigen function Y such that derivative 
of Y does not changes its sign. 
Letk-,. 
Then, or 
( • every constant multiple of y(x) is also 
eigen function) 
Zeroes of eigen functions are simple 
(multiplicity 1) 
(c) is incorrect


Machine generated alternative text:
(25) 
and where and 
are continuous and 
on then Let 
where -Ek, are eigen space 
corresponding to eigen value then 
dim Ex, 
Countably infinite 
-2 for all except one value of


Machine generated alternative text:
Answer : 
Explanation : 
is least eigen value as corresponding 
eigen function does not change its sign. 
Also dim A (•:r.u1arSLP) 
And vector space over same field with 
same dimension are similar.

Machine generated alternative text:
Answer : 
Explanation : 
This is periodic SLP and by its 
properties a, b, c, d are true

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