Q15) Assume that the roots of a characteristic polynomial of a homogeneous ODE with constant coefficients are 0,0,0,2+51,2-51, then the general solutiom of this ODE, is: ₂x² + e2x [Acos(x) + Bsin(x)] a) c₂ + ₂x + b) c₂ + ₂x + c) ₂x + ₂x² ₂x² +ex (Acos(2x) + Bsin(2x)] + ₂x³ + e*[Acos(5x) + Bsin(5x)] d) G₂ +G₂x + ₂x² + ²x[Acos(5x) + Bsin(5x)]

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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SOLVE Q15
Q10) The general solution of y" +2y" -y'-2y = 0, is:
a) y(x) = c₂e²x+c₂e-² +₂ex
c) y(x) = c₂e²x + c₂e*
b) y(x) = c₂e-2x + c₂ex
Q11) L(y") =
d) y(x) = c₂e-²x + c₂e²+G₂e™*
a) s² L(y(x)) + sy(0) - y'(0)
c) s² L(y(x)) -sy(0)-y'(0)
Q12) The kernel of the Laplace transform
b) e-t
b) s² L(y(x))+ sy(0) + y'(0)
of f(t),t> 0, is:
d) s² L(y(x))-sy(0) + y'(0)
c) est
Q13) Evaluate (e-t sin4t):
a)
842
b)
342
x² +8+20
d) +20
Q14) If the power series method was used to solve the following ODE.
(x-0.5)y"
x+0.5
y' + (x²-1) y = 0.xo = 0. Then the interval of convergence, is:
b) (-1,1)
c) (0,00)
(9)
Q15) Assume that the roots of a characteristic polynomial of a homogeneous ODE with constant coefficients are
0,0,0,2+51,2-51, then the general solutiom of this ODE, is:
a) c₂ + ₂x + ₂x² + e2x [Acos(x) + Bsin(x)]
b) c₂ + ₂x + ₂x² +e5x (Acos(2x) + Bsin(2x)]
c) c₂x + ₂x² + 3x³ + e*[Acos(5x) + Bsin(5x)]
d) G₂ + G₂x + C₂x² + ²x [Acos(5x) + Bsin(5x)]
Q16) Combine the following power series expressions into a single power series.
Σ(n+1)(x-2)-1 +Eon(x - 2)"
a) (2n + 1)(x)"
b) To(2n + 2)(x)"
c) Eno(2n + 2)(x - 2)"
d)
(2n+1)(x-2)"
Q17) If you know that the radius of convergent of the series method for the ODE
y"+y+=0, x=2 is 5. Find the value for b, (b<0):
(a)-5
(b)-3
(c)-2
-8
Q18) The formula of the particular solution yp of y) + 4y" = 3 sin(2t) - 5cos2t, is:
a) yp= Asin(t) + Bcos(t)
b) yp
Atsin(t) + Btcos(t)
.
c) y₂ = Asin(2t) + Bcos(2t)
d) y = Atsin(2t) + Btcos(2t)
Q19) est dt =
a)
b) 0
9)
d) ∞o
Q20) The singular point (s) of (x + 1) y' + x²y = 0, is (are):
a) 0,1
b) 0,-1
c) 0
d)-1
1
=
b) 1-cosht c) (cosh2t - 1) d) (1-cosh2t)
2t2y" +ty' - 3y = 0,t> 0, then y₂ (t) is:
Q21) L1
a) cosht - 1
s(s²-1)
Q22) Given that y₁ (t) = t¹ is solution for
a)t b) ti
Q23) L-1
c) ti
d) ta
estsinst
-atsinst
1
a)
b)
est sin3t
5
c)
e-at sin3t
5
d)
5
5
(s-3)2+25A
Q24) The general solution for y' = 6y²x, is:
a) == 3x² + c
b) ² = x² + c
c) =² = 3x² + c
d) == x² + c
y
Q25) The form of a particular solution of y"-4y' - 12y = sin(2t), is:
a) y(t) = A sin(2t)
b) y(t) = A cos(t)
c) y(t) = A cos(2t) + B sin(2t)
d) y(t) = A cos (t) + B sin(t)
Transcribed Image Text:Q10) The general solution of y" +2y" -y'-2y = 0, is: a) y(x) = c₂e²x+c₂e-² +₂ex c) y(x) = c₂e²x + c₂e* b) y(x) = c₂e-2x + c₂ex Q11) L(y") = d) y(x) = c₂e-²x + c₂e²+G₂e™* a) s² L(y(x)) + sy(0) - y'(0) c) s² L(y(x)) -sy(0)-y'(0) Q12) The kernel of the Laplace transform b) e-t b) s² L(y(x))+ sy(0) + y'(0) of f(t),t> 0, is: d) s² L(y(x))-sy(0) + y'(0) c) est Q13) Evaluate (e-t sin4t): a) 842 b) 342 x² +8+20 d) +20 Q14) If the power series method was used to solve the following ODE. (x-0.5)y" x+0.5 y' + (x²-1) y = 0.xo = 0. Then the interval of convergence, is: b) (-1,1) c) (0,00) (9) Q15) Assume that the roots of a characteristic polynomial of a homogeneous ODE with constant coefficients are 0,0,0,2+51,2-51, then the general solutiom of this ODE, is: a) c₂ + ₂x + ₂x² + e2x [Acos(x) + Bsin(x)] b) c₂ + ₂x + ₂x² +e5x (Acos(2x) + Bsin(2x)] c) c₂x + ₂x² + 3x³ + e*[Acos(5x) + Bsin(5x)] d) G₂ + G₂x + C₂x² + ²x [Acos(5x) + Bsin(5x)] Q16) Combine the following power series expressions into a single power series. Σ(n+1)(x-2)-1 +Eon(x - 2)" a) (2n + 1)(x)" b) To(2n + 2)(x)" c) Eno(2n + 2)(x - 2)" d) (2n+1)(x-2)" Q17) If you know that the radius of convergent of the series method for the ODE y"+y+=0, x=2 is 5. Find the value for b, (b<0): (a)-5 (b)-3 (c)-2 -8 Q18) The formula of the particular solution yp of y) + 4y" = 3 sin(2t) - 5cos2t, is: a) yp= Asin(t) + Bcos(t) b) yp Atsin(t) + Btcos(t) . c) y₂ = Asin(2t) + Bcos(2t) d) y = Atsin(2t) + Btcos(2t) Q19) est dt = a) b) 0 9) d) ∞o Q20) The singular point (s) of (x + 1) y' + x²y = 0, is (are): a) 0,1 b) 0,-1 c) 0 d)-1 1 = b) 1-cosht c) (cosh2t - 1) d) (1-cosh2t) 2t2y" +ty' - 3y = 0,t> 0, then y₂ (t) is: Q21) L1 a) cosht - 1 s(s²-1) Q22) Given that y₁ (t) = t¹ is solution for a)t b) ti Q23) L-1 c) ti d) ta estsinst -atsinst 1 a) b) est sin3t 5 c) e-at sin3t 5 d) 5 5 (s-3)2+25A Q24) The general solution for y' = 6y²x, is: a) == 3x² + c b) ² = x² + c c) =² = 3x² + c d) == x² + c y Q25) The form of a particular solution of y"-4y' - 12y = sin(2t), is: a) y(t) = A sin(2t) b) y(t) = A cos(t) c) y(t) = A cos(2t) + B sin(2t) d) y(t) = A cos (t) + B sin(t)
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