3.-. 3) The general solution of y" - 5y' +6y=0 is. A. y = ce² +₂e³ B. y ce2 + c₂e-3 C. y cie-2z+c₂e³ D. y ce-2 + c₂e-3z

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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Q 3 please
1.-. e) The solution of = 5, y(0) = 1 is.
A. x² = y² + 1
B. y² = x² + 1
2. (
A.
3.-
5. (.
8.
A. y = c₁e² +₂e³
4. -
s) The general solution of r2y" + 3xy' + y = 0 is.
A. y = ₁x + ₂x² B. y c₁x + c₂xln(x)
9.
10.
s) The integrating factor of xy' + 3y = is.
B. 2
C. 2³
A. 3
7. () The value of L{} is.
A. e-t
B. et
11.
3) The general solution of y" - 5y' +6y=0 is.
B. y=c₁e²+ c₂e-32
C. 00
D. 3
6.-
s) The minimum radius of convergence of power sreies solution about z = 1 of the ODE
(x2 - 49)y" + 2xy' + y = 0..
B. 8
A. 7
A. 1
5) The radius of convergence of E(-1)*(-1)
B. 0
8+3
The value of L-¹22) is.
82+28+1
B. 2tet + et
A. 3tet + e-t
The value of Lu(t-3)} is.
B. e-(-+)
A. e³a(+2)
-
C. y² = 2x² + 1 D. y² - x² = 0.
The equivalent form of
Using unit step functions is
A. u(t-2) (sin(t)-3) +3
D. 3u(t-2) - sin(t)u(t-2)
f(t) =
C. y=c₁e-2z+c₂e³z
C. y = 9+ln(z)
The value of f sin(t)8(t-1)dt
B. 0
C. 6
C. u(t)
= { sin()
C. 2te-t + e-t
C. e³8 (++)
0<t<2
sin(t) t>2
B. 3u(t-2) + sin(t)u(t - 2)
D.
C.
D. y cie-2 + c₂e-3z
D. y=+
D. 5
D. 8(t)
D. 3tet + et
D. e38 (++)
C. u(t-2) (sin(t) + 3) +3
D. √2
Transcribed Image Text:1.-. e) The solution of = 5, y(0) = 1 is. A. x² = y² + 1 B. y² = x² + 1 2. ( A. 3.- 5. (. 8. A. y = c₁e² +₂e³ 4. - s) The general solution of r2y" + 3xy' + y = 0 is. A. y = ₁x + ₂x² B. y c₁x + c₂xln(x) 9. 10. s) The integrating factor of xy' + 3y = is. B. 2 C. 2³ A. 3 7. () The value of L{} is. A. e-t B. et 11. 3) The general solution of y" - 5y' +6y=0 is. B. y=c₁e²+ c₂e-32 C. 00 D. 3 6.- s) The minimum radius of convergence of power sreies solution about z = 1 of the ODE (x2 - 49)y" + 2xy' + y = 0.. B. 8 A. 7 A. 1 5) The radius of convergence of E(-1)*(-1) B. 0 8+3 The value of L-¹22) is. 82+28+1 B. 2tet + et A. 3tet + e-t The value of Lu(t-3)} is. B. e-(-+) A. e³a(+2) - C. y² = 2x² + 1 D. y² - x² = 0. The equivalent form of Using unit step functions is A. u(t-2) (sin(t)-3) +3 D. 3u(t-2) - sin(t)u(t-2) f(t) = C. y=c₁e-2z+c₂e³z C. y = 9+ln(z) The value of f sin(t)8(t-1)dt B. 0 C. 6 C. u(t) = { sin() C. 2te-t + e-t C. e³8 (++) 0<t<2 sin(t) t>2 B. 3u(t-2) + sin(t)u(t - 2) D. C. D. y cie-2 + c₂e-3z D. y=+ D. 5 D. 8(t) D. 3tet + et D. e38 (++) C. u(t-2) (sin(t) + 3) +3 D. √2
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