Q1: If y" + y = x², then yp = Q4: The solution of y(4) - 5y" + 4y = 0 is: (B) c₁e* + c₂ex + c3e²x+c4e-2x (A) x² Q2: If y" + 3y' +2.25y = -10e-1.5x, then yp = (A) Ax²e-15x (B) Ax²e-1.5x + Bxe-1.5x C) Ax²e-1.5x + Bxe + Ce-1.5x Q3: The Wronskian of the homogeneous solution of y" + y = cscx, is= (A) -1 (C) c₁e* + c₂e4x Q5: The characteristic equation of x³y""-3x²y" + 6xy' -6y = 0, is: (B) m³ + 6m² -11m+6=0 Q6: One of the following is linearly dependent set. (A) {x², x, 1} Q7: The solution of y(4) = 0, is: (A) Cx³ (B) C₁+C₂x + C3x² + C4x³ (C) C₁+C₂ex + С3e²x + С₁e³x (A) an n+2 (B) x² + 2 , n = 1,2,... Q10: The solution of y" + xy = 0, is y = (A) a₁ + a₁x + ªx² + .... 90x3 (B) ao + a₁x + 6 1.5) (C) m³ - 6m² + 11m + 6 = 0(D) m³-6m² -11m-6=0 (B) {x², 1, 1) (C) (ex.x, 1} (D) Cısinx + C2cosx + Q8: The recurrence relation of the power series solution of y" + xy + y = 0, is a₂ = 20, and a₂+2 = (B) (n-1) an (n+1)(n+2) (C)=n= 1,2,... ,n=1,2,... (D) n = 1,2,... (B) 2 a0x3 12 Q9: Use L{1}={coswt} = 2+2 to find L-¹, ¹ { ₁ ³ + 45-)} = (A) ³ S Q11: Use cos2x = 2cos²x - 1and £{1}=and L{coswx}= (A) = (-²3-52464) (B) =/= ( = +5² +64) Q12: If y"= 1, then yp= (A) x² 1-cos2t 4 (C) x² - 2 (C) ao + a₁x-. - (D) Ae-1.5x (B) (C) 0 (D) 1 (A) C₁e* + C₂xe* + c3e²x+₁е-2x (D) c₁e²x + c₂e-2x + C3sinx + C4cOSX (A) m³-6m² + 11m-6=0 (B) x² + x + 1 S s²+w² (C) =/= ( ² - 5 ² + 654) 90x³ 12 (1-n)an (n+1)(n+2)' (1-cos3t) (D) 2x² (C) to find L{cos2 (4x)} = (D) {x².0) 1+cos (D) ao + a₁x- sinx +C,xco: (D) 2 (1+cos3t) apx³ 6 (0) ²/2 ( ²3 +52² 4/64) +... (D) x + 1
Q1: If y" + y = x², then yp = Q4: The solution of y(4) - 5y" + 4y = 0 is: (B) c₁e* + c₂ex + c3e²x+c4e-2x (A) x² Q2: If y" + 3y' +2.25y = -10e-1.5x, then yp = (A) Ax²e-15x (B) Ax²e-1.5x + Bxe-1.5x C) Ax²e-1.5x + Bxe + Ce-1.5x Q3: The Wronskian of the homogeneous solution of y" + y = cscx, is= (A) -1 (C) c₁e* + c₂e4x Q5: The characteristic equation of x³y""-3x²y" + 6xy' -6y = 0, is: (B) m³ + 6m² -11m+6=0 Q6: One of the following is linearly dependent set. (A) {x², x, 1} Q7: The solution of y(4) = 0, is: (A) Cx³ (B) C₁+C₂x + C3x² + C4x³ (C) C₁+C₂ex + С3e²x + С₁e³x (A) an n+2 (B) x² + 2 , n = 1,2,... Q10: The solution of y" + xy = 0, is y = (A) a₁ + a₁x + ªx² + .... 90x3 (B) ao + a₁x + 6 1.5) (C) m³ - 6m² + 11m + 6 = 0(D) m³-6m² -11m-6=0 (B) {x², 1, 1) (C) (ex.x, 1} (D) Cısinx + C2cosx + Q8: The recurrence relation of the power series solution of y" + xy + y = 0, is a₂ = 20, and a₂+2 = (B) (n-1) an (n+1)(n+2) (C)=n= 1,2,... ,n=1,2,... (D) n = 1,2,... (B) 2 a0x3 12 Q9: Use L{1}={coswt} = 2+2 to find L-¹, ¹ { ₁ ³ + 45-)} = (A) ³ S Q11: Use cos2x = 2cos²x - 1and £{1}=and L{coswx}= (A) = (-²3-52464) (B) =/= ( = +5² +64) Q12: If y"= 1, then yp= (A) x² 1-cos2t 4 (C) x² - 2 (C) ao + a₁x-. - (D) Ae-1.5x (B) (C) 0 (D) 1 (A) C₁e* + C₂xe* + c3e²x+₁е-2x (D) c₁e²x + c₂e-2x + C3sinx + C4cOSX (A) m³-6m² + 11m-6=0 (B) x² + x + 1 S s²+w² (C) =/= ( ² - 5 ² + 654) 90x³ 12 (1-n)an (n+1)(n+2)' (1-cos3t) (D) 2x² (C) to find L{cos2 (4x)} = (D) {x².0) 1+cos (D) ao + a₁x- sinx +C,xco: (D) 2 (1+cos3t) apx³ 6 (0) ²/2 ( ²3 +52² 4/64) +... (D) x + 1
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
please solve first two questions
![Q1: If y" + y = x², then yp =
(A) x²
Q2: If y" + 3y' +2.25y = -10e-1.5x, then yp =
(A) Ax²e-15x (B) Ax²e-1.5x + Bxe-1.5x C) Ax²e-1.5x + Bxe-1.5x + Ce-1.5x (D) Ae-1.5x
Q3: The Wronskian of the homogeneous solution of y" + y = cscx, is= (A)-1
Q4: The solution of y(4) - 5y" + 4y = 0 is:
(B) c₁e* + c₂ex + c3e²x + ₁е-2x
(C) c₁e* + c₂e4x
Q5: The characteristic equation of x³y" - 3x²y" + 6xy' - 6y = 0, is:
(B) m³ + 6m² -11m+6=0
Q6: One of the following is linearly dependent set. (A) {x², x, 1}
Q7: The solution of y(4) = 0, is:
(A) Cx³ (B) C₁ + C₂x + C3x² + 4x³ (C) C₁+C₂ex + С3e²x + ₁e³x
n+2'
n = 1,2,...
S
Q9: Use L{1}={coswt} = 52+2 to find L-1,
Q10: The solution of y" + xy = 0, is y =
(B) ao + a₁x +
aox³
6
(A) ao + a₁x +
·+...
Q8: The recurrence relation of the power series solution of y" + xy + y = 0, is a₂ = 20, and a₂+2 =
(A).
(B)
(n-1) an
(n+1)(n+2)
n = 1,2,...
(C)2, n = 1,2,...
(D)
n = 1,2,...
(B) x² + 2
(C) m³ - 6m² + 11m + 6 = 0(D) m³-6m² -11m-6=0
(B) {x², 1,1}
(C) (ex.x, 1}
aox³
12
Q11: Use cos2x = 2cos²x - 1and L{1}=and
(A) = (²-²-5² +64)
(B) / (+52 +64)
Q12: If y"= 1, then yp =
(A) x²
1
¹ { ₁ ³ + 45 -} = (A)
+4s
+...
(C) x²-2
1-cos2t
4
(C) 0
(D) 1
(A) ce* +Czxe* +c3e2x +ce-2x
(D) c₁e²x + c₂e-2x + c3sinx + C4cosx
(A) m³ - 6m² + 11m-6=0
(B) x² + x + 1
(D) Cısinx + C2cosx +Cgxsinx +Caxco:
(1-n)an
(n+1)(n+2)'
(D) 2x²
(B) 2 (1-cost) (C)
9
(C) ao + a₁x--
S
L{coswx} = 52+2 to find L{cos² (4x)} =
(C) =/= ( 3 - 5 ² ²64)
40x3
12
(D) {x², 112,0}
(C)=x²
1+cos
4
(D) 2 (1+cos3t)
(D) ao + a₁x.
aox³
6
(D) ²/2 ( = +52 4/64)
+...
(D) x + 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6786f013-643c-4548-9b80-003a487634b7%2F6a625dd6-82f7-4bab-8599-0025a10ebc49%2Ffjn4gam_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q1: If y" + y = x², then yp =
(A) x²
Q2: If y" + 3y' +2.25y = -10e-1.5x, then yp =
(A) Ax²e-15x (B) Ax²e-1.5x + Bxe-1.5x C) Ax²e-1.5x + Bxe-1.5x + Ce-1.5x (D) Ae-1.5x
Q3: The Wronskian of the homogeneous solution of y" + y = cscx, is= (A)-1
Q4: The solution of y(4) - 5y" + 4y = 0 is:
(B) c₁e* + c₂ex + c3e²x + ₁е-2x
(C) c₁e* + c₂e4x
Q5: The characteristic equation of x³y" - 3x²y" + 6xy' - 6y = 0, is:
(B) m³ + 6m² -11m+6=0
Q6: One of the following is linearly dependent set. (A) {x², x, 1}
Q7: The solution of y(4) = 0, is:
(A) Cx³ (B) C₁ + C₂x + C3x² + 4x³ (C) C₁+C₂ex + С3e²x + ₁e³x
n+2'
n = 1,2,...
S
Q9: Use L{1}={coswt} = 52+2 to find L-1,
Q10: The solution of y" + xy = 0, is y =
(B) ao + a₁x +
aox³
6
(A) ao + a₁x +
·+...
Q8: The recurrence relation of the power series solution of y" + xy + y = 0, is a₂ = 20, and a₂+2 =
(A).
(B)
(n-1) an
(n+1)(n+2)
n = 1,2,...
(C)2, n = 1,2,...
(D)
n = 1,2,...
(B) x² + 2
(C) m³ - 6m² + 11m + 6 = 0(D) m³-6m² -11m-6=0
(B) {x², 1,1}
(C) (ex.x, 1}
aox³
12
Q11: Use cos2x = 2cos²x - 1and L{1}=and
(A) = (²-²-5² +64)
(B) / (+52 +64)
Q12: If y"= 1, then yp =
(A) x²
1
¹ { ₁ ³ + 45 -} = (A)
+4s
+...
(C) x²-2
1-cos2t
4
(C) 0
(D) 1
(A) ce* +Czxe* +c3e2x +ce-2x
(D) c₁e²x + c₂e-2x + c3sinx + C4cosx
(A) m³ - 6m² + 11m-6=0
(B) x² + x + 1
(D) Cısinx + C2cosx +Cgxsinx +Caxco:
(1-n)an
(n+1)(n+2)'
(D) 2x²
(B) 2 (1-cost) (C)
9
(C) ao + a₁x--
S
L{coswx} = 52+2 to find L{cos² (4x)} =
(C) =/= ( 3 - 5 ² ²64)
40x3
12
(D) {x², 112,0}
(C)=x²
1+cos
4
(D) 2 (1+cos3t)
(D) ao + a₁x.
aox³
6
(D) ²/2 ( = +52 4/64)
+...
(D) x + 1
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