a) f(x), g(x) and h(x) are c) W (f(x), g(x), h(x)) = 0 d) W (f 02) 6-1 (22) H a)-2coshnt b)-2sinhnt c)-2cosnt d)-2sinnt Q3) The Integrating factor which make (3x²y + 2xy + y³)dx + (x² + y²)dy = 0 exact, is: a) e-3x b) ex c) ex d) ex Q4) The linear form of nonlinear ODE y' - 2y = 2y, is: a) u' + 6u = -6 b) u'-6u = -6 c) u' - 6u = 6 d) u' + 6u = 6 Q5) The general solution of 2x2y" + 3xy' - 15y = 0, is: a) y(x) = c₂x² + ₂x³ b) y(x) = ₂x + ₂x-3 S ©)x(x) = GX+G d) y(x) = ₂x + ₂x³ -0.5 cos3t): Q6) Evaluate L (2e-2t sin4t 8 8 a) b) (-2)³+16 S 2s² +18 (s+2)+16 c)(8+2)³+16 25³ +18 25² +18 (s-2)³ +16 b)y(t) = cet cos(√5 t) + c₂e²sin (√5t) d) y(t) = Gel cos(5 t) + c₂e²sin (5 t) Q7) The general solution of y"-4y' +9y = 0, is: a) y(t) = c₁e²t cos(5 t) + c₂e²t sin (5 t) c)y(t) = c₂e²t cos(√5 t) + c₂e2tsin (√5 t) Q8) The inverse Laplace transform of H(s) +²e² c) f(t) ==e+e-2t is: * (3+2)(x-2) -2t a) f(t) == b) f(t) ==e+ d)f(t) == +²t Q9) The solution of y"+y' = 0 by using power series method, is: *** a) y(x) = a + a₁ (1- .) 31 41 St 24 c) y(x) = a + a₁(x-+ 41 51 2s²+18 2/52/5 + b) y(x) = a + a₂(1+-+- d)y(x) = a + a₂(x + - + +)
a) f(x), g(x) and h(x) are c) W (f(x), g(x), h(x)) = 0 d) W (f 02) 6-1 (22) H a)-2coshnt b)-2sinhnt c)-2cosnt d)-2sinnt Q3) The Integrating factor which make (3x²y + 2xy + y³)dx + (x² + y²)dy = 0 exact, is: a) e-3x b) ex c) ex d) ex Q4) The linear form of nonlinear ODE y' - 2y = 2y, is: a) u' + 6u = -6 b) u'-6u = -6 c) u' - 6u = 6 d) u' + 6u = 6 Q5) The general solution of 2x2y" + 3xy' - 15y = 0, is: a) y(x) = c₂x² + ₂x³ b) y(x) = ₂x + ₂x-3 S ©)x(x) = GX+G d) y(x) = ₂x + ₂x³ -0.5 cos3t): Q6) Evaluate L (2e-2t sin4t 8 8 a) b) (-2)³+16 S 2s² +18 (s+2)+16 c)(8+2)³+16 25³ +18 25² +18 (s-2)³ +16 b)y(t) = cet cos(√5 t) + c₂e²sin (√5t) d) y(t) = Gel cos(5 t) + c₂e²sin (5 t) Q7) The general solution of y"-4y' +9y = 0, is: a) y(t) = c₁e²t cos(5 t) + c₂e²t sin (5 t) c)y(t) = c₂e²t cos(√5 t) + c₂e2tsin (√5 t) Q8) The inverse Laplace transform of H(s) +²e² c) f(t) ==e+e-2t is: * (3+2)(x-2) -2t a) f(t) == b) f(t) ==e+ d)f(t) == +²t Q9) The solution of y"+y' = 0 by using power series method, is: *** a) y(x) = a + a₁ (1- .) 31 41 St 24 c) y(x) = a + a₁(x-+ 41 51 2s²+18 2/52/5 + b) y(x) = a + a₂(1+-+- d)y(x) = a + a₂(x + - + +)
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 Q9
![Q1) One of the following is correct for functions the f(x) = 1, g(x)= x³ and h(x)
M
a) f(x), g(x) and h(x) are linearly dependent
c) W (f(x), g(x), h(x)) = 0
b) f(x), g(x) and h(x) are linearly independent
d) W(f(x), g(x), h(x)) = 9
Q2) L-1
-2s
2+3
H
a) -2coshnt
b)-2sinhnt
c)-2cosnt
d)-2sinnt
Q3) The Integrating factor which make (3x2y + 2xy + y³)dx + (x² + y²)dy = 0 exact, is:
a) e-3x
b) ei*
d) e³x
Q4) The linear form of nonlinear ODE y' - 2y = 2y, is:
a) u' + 6u = -6
b) u' - 6u = -6
c) u'-6u = 6
d) u' + 6u = 6
Q5) The general solution of 2x2y" + 3xy'-15y = 0, is:
a) y(x) = ₂x² + ₂x³
b) y(x) = ₂x + ₂x-3
c) y(x) = c₁x² + ₂x-3
d) y(x) = c₁x + ₂x³
Q6) Evaluate L (2 e-2 sin4t -0.5 cos3t):
8
S
a)
(s+2)² +16
b)
c) (8+2)² +16 23²418
(5-2)2+16
25² +18
2s² +18
(s-2)² +16
Q7) The general solution of y"-4y' +9y = 0, is:
a) y(t)=ce²t cos(5 t) + c₂e²tsin (5 t)
c)y(t) = cet cos(√5 t) + c₂e²t sin (√5 t)
Q8) The inverse Laplace transform of H(s):
b)y(t) = ce cos(√5 t) + c₂e¹sin (√5t)
d) y(t) = ce cos(5 t) + cesin (5 t)
1
=
is:
.
(35+2)(5-2)
a) f(t) ==e+e²t
b) f(t) ==e.
8
c) f(t) ==e+e-2t
d)f(t) ==e.
+²e²t
Q9) The solution of y" + y'= 0 by using power series method, is:
a) y(x) = ao + a₁ (1-
11
b) y(x) = a + a₁(1+
21
31
41
St
21
c) y(x) = a₁ + a₁(x-+
d)y(x) = a + a₂(x + ==
51
-
e-2t
1
2s² +18
+ +
ii
41
41
51
++
+ ...)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc7df285d-c30a-4214-bfbd-28e975c7e2db%2F3c4a2f03-3ca2-496d-a4cd-25d8abe27fe0%2Fiqczrwt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q1) One of the following is correct for functions the f(x) = 1, g(x)= x³ and h(x)
M
a) f(x), g(x) and h(x) are linearly dependent
c) W (f(x), g(x), h(x)) = 0
b) f(x), g(x) and h(x) are linearly independent
d) W(f(x), g(x), h(x)) = 9
Q2) L-1
-2s
2+3
H
a) -2coshnt
b)-2sinhnt
c)-2cosnt
d)-2sinnt
Q3) The Integrating factor which make (3x2y + 2xy + y³)dx + (x² + y²)dy = 0 exact, is:
a) e-3x
b) ei*
d) e³x
Q4) The linear form of nonlinear ODE y' - 2y = 2y, is:
a) u' + 6u = -6
b) u' - 6u = -6
c) u'-6u = 6
d) u' + 6u = 6
Q5) The general solution of 2x2y" + 3xy'-15y = 0, is:
a) y(x) = ₂x² + ₂x³
b) y(x) = ₂x + ₂x-3
c) y(x) = c₁x² + ₂x-3
d) y(x) = c₁x + ₂x³
Q6) Evaluate L (2 e-2 sin4t -0.5 cos3t):
8
S
a)
(s+2)² +16
b)
c) (8+2)² +16 23²418
(5-2)2+16
25² +18
2s² +18
(s-2)² +16
Q7) The general solution of y"-4y' +9y = 0, is:
a) y(t)=ce²t cos(5 t) + c₂e²tsin (5 t)
c)y(t) = cet cos(√5 t) + c₂e²t sin (√5 t)
Q8) The inverse Laplace transform of H(s):
b)y(t) = ce cos(√5 t) + c₂e¹sin (√5t)
d) y(t) = ce cos(5 t) + cesin (5 t)
1
=
is:
.
(35+2)(5-2)
a) f(t) ==e+e²t
b) f(t) ==e.
8
c) f(t) ==e+e-2t
d)f(t) ==e.
+²e²t
Q9) The solution of y" + y'= 0 by using power series method, is:
a) y(x) = ao + a₁ (1-
11
b) y(x) = a + a₁(1+
21
31
41
St
21
c) y(x) = a₁ + a₁(x-+
d)y(x) = a + a₂(x + ==
51
-
e-2t
1
2s² +18
+ +
ii
41
41
51
++
+ ...)
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