- am-1 = (m + 2)(m + 1)a,m+2 %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Solve Q10
![Q.1) The general solution of the equation y* • v• ' -e' cos x , ÁN:
a) y(x ) =c, +c; x +c,¢ ' Ae' cosa a Re' sin x
b) y (x ) =c, +c,x +c,¢
+ de' cos a + Be' sinx
c) y(x) = Ce¯' + ei(Acosx + Bsinx) + xe
d) v (x ) -c, +c, x +c,e¨ +Ae ' cosx + Be ' sin x
Q.2) The basis of the equation y'' + y" = 0 ,are:
a) 1,1.e
Q.3) Ihe radius of convergence of (1- x²)y" + 4y' + 6xy = 0 about xo = 0 , is:
b) 1 , 1, e¯'
c) 1,x,e²
d) 1 , x , e"
%3D
a)+ o
b) 2
c) 0
b) - 12e?* с) бе?
d) 1
d) -6e"
Q.4) The Wronskian W(e~2", e", e²*), is:
Q.5) The general solution of the equation y" + 9y" = 0 , is:
a) y(x) = c; + cz + Acos2x + Bsin2x
c) y(x) = c; + c; + Acos3x + Bsin3x
Q.6) The singular point (s) of (x² + 4)y" – x y' + 4y = 0 , is (arc):
a) 12
Q.7) To write the power series Em=3 4m as the form £-zum ,we must shift m to become :
a) m - 1
Q.8) The characteristic equation for y"' + 3y" – 4y = 0 , is :
a) 2' = 32?
Q.9) The form of the particular solution of the equation y" – 9y = 4 sin 3x + 2cos3x , is:
a) y,(x) = A cos3x
c) y, (x) = A sin3x
Q.10) The recurrence relation which obtained from solving y" + xy = 0 by using power series , is:
a) 12e*
b)y(x) = c, + c,x + Acos2x + Bsin2x
d) y(x) = c, + c,x + Acos3x + Bsin3x
b) ±2i
c) 0
d) (0 , ±2i }
b) т - 2
с) т +1
d) m +2
b) 23 = 322-4
c) d* + 2 = -1
d) 2' = -32? + 4
%3D
b) y,(x) = A cos3x + Bsin3x
d) y, (x) = Axcos3x + Bxsin3x
a) – am-1 = (m + 2)(m + 1)a,m+2
b) am+2
%3D
(m +2)(m+1)
с) а, 3D (т — 2)(m - 1)а,m+2
d) am+2
(m+2)(m+ 1)
Q.11) Rewrite the series x E=1m am x™-1 + £m-0 amx™ as a single power series whose general term
involves xm:
a) E=s(am-1 – am-4) x™
c) a, + Em-5(am-1- am-4) x™
Q.12) The solution of y" + y' = 0 by using power series method , is:
b) E-4(am-1 – am-3) x™
d) a, + Em=1(m am + am) x™
Am-4) xm
a) y(x) = ao – a;(1 –
b) y(x) = a, + a(1+
2!
3!
x*
с) у (x) %3D а, + а,(х-:
d)y(x) = ao + a,( x +
+ ..)
+
5!
2!
4!
(-1)™x"
Q.13) The interval and radius of convergence for power series Em-o
b) (-∞, +m), 1
, respectively , are :
d) (-m, +∞), 3
m!
a) (-∞, +∞), ∞
c) (-∞, +∞), 2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd924b0f1-2d9d-4acf-96cb-172e6fe21980%2Fd39524cd-8702-46ba-aaa9-147a18949145%2Fxkuz8qf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q.1) The general solution of the equation y* • v• ' -e' cos x , ÁN:
a) y(x ) =c, +c; x +c,¢ ' Ae' cosa a Re' sin x
b) y (x ) =c, +c,x +c,¢
+ de' cos a + Be' sinx
c) y(x) = Ce¯' + ei(Acosx + Bsinx) + xe
d) v (x ) -c, +c, x +c,e¨ +Ae ' cosx + Be ' sin x
Q.2) The basis of the equation y'' + y" = 0 ,are:
a) 1,1.e
Q.3) Ihe radius of convergence of (1- x²)y" + 4y' + 6xy = 0 about xo = 0 , is:
b) 1 , 1, e¯'
c) 1,x,e²
d) 1 , x , e"
%3D
a)+ o
b) 2
c) 0
b) - 12e?* с) бе?
d) 1
d) -6e"
Q.4) The Wronskian W(e~2", e", e²*), is:
Q.5) The general solution of the equation y" + 9y" = 0 , is:
a) y(x) = c; + cz + Acos2x + Bsin2x
c) y(x) = c; + c; + Acos3x + Bsin3x
Q.6) The singular point (s) of (x² + 4)y" – x y' + 4y = 0 , is (arc):
a) 12
Q.7) To write the power series Em=3 4m as the form £-zum ,we must shift m to become :
a) m - 1
Q.8) The characteristic equation for y"' + 3y" – 4y = 0 , is :
a) 2' = 32?
Q.9) The form of the particular solution of the equation y" – 9y = 4 sin 3x + 2cos3x , is:
a) y,(x) = A cos3x
c) y, (x) = A sin3x
Q.10) The recurrence relation which obtained from solving y" + xy = 0 by using power series , is:
a) 12e*
b)y(x) = c, + c,x + Acos2x + Bsin2x
d) y(x) = c, + c,x + Acos3x + Bsin3x
b) ±2i
c) 0
d) (0 , ±2i }
b) т - 2
с) т +1
d) m +2
b) 23 = 322-4
c) d* + 2 = -1
d) 2' = -32? + 4
%3D
b) y,(x) = A cos3x + Bsin3x
d) y, (x) = Axcos3x + Bxsin3x
a) – am-1 = (m + 2)(m + 1)a,m+2
b) am+2
%3D
(m +2)(m+1)
с) а, 3D (т — 2)(m - 1)а,m+2
d) am+2
(m+2)(m+ 1)
Q.11) Rewrite the series x E=1m am x™-1 + £m-0 amx™ as a single power series whose general term
involves xm:
a) E=s(am-1 – am-4) x™
c) a, + Em-5(am-1- am-4) x™
Q.12) The solution of y" + y' = 0 by using power series method , is:
b) E-4(am-1 – am-3) x™
d) a, + Em=1(m am + am) x™
Am-4) xm
a) y(x) = ao – a;(1 –
b) y(x) = a, + a(1+
2!
3!
x*
с) у (x) %3D а, + а,(х-:
d)y(x) = ao + a,( x +
+ ..)
+
5!
2!
4!
(-1)™x"
Q.13) The interval and radius of convergence for power series Em-o
b) (-∞, +m), 1
, respectively , are :
d) (-m, +∞), 3
m!
a) (-∞, +∞), ∞
c) (-∞, +∞), 2
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