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3) The solution of, y(1) - 1 is. B. y²=²+1 A. r² = y² +1 C. y² = 2x² +1 D. ²-²=0.

3) The solution of, y(1) - 1 is. B. y²=²+1 A. r² = y² +1 C. y² = 2x² +1 D. ²-²=0.

Advanced Engineering Mathematics
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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A. x² = y² +1 3. A. 4. ( 5. A. y cie + c₂e6 3) The solution of = 5, y(1) = 1 is. B. y2x2+1 A. 8. A. y = ₁x + ₂x² us) The general solution of y" + 5y' -6y = 0 is. B. y cie-6 + c₂e 9. ( The integrating factor of xy' + 2y = is. C. ³ B. 2 A. e-t -) The general solution of r2y" - 3xy' + 4y = 0 is. B. y cr² + c₂r²ln(r) The radius of convergence of 1(-1)*3kk(x-4). B. 0 C. 00 11. ( A. 3tet + et D. 3 6. (s) The minimum radius of convergence of power series solution about r= -2 of the ODE (x²-49)y" + 2xy' + y = 0.. A. 7 B. 8 s) The value of {} is. B. et 844 3) The value of L-1{} is. B. 2tet + et A. 1 A. e²(+²) - C. y² = 2x² +1 The value of L{u(t-2)} is. B. e²(++) 10. (The equivalent form of Using unit step functions is A. u(t-3) (tan(t) + 4) + 4 D. 4u(t-3)-tan(t)u(t-3) C. ycie + c₂e-bx f(t) = C. y=+(2) C. 6 The value of f tan(t)o(t-)dt B. 0 C. u(t) C. 2te-t + et D. ²-x²=0. 0<t<3 D. D. y cez + cela C. ₂ D. y = 2+2 D. 5 C. e(++) D. e² (3+²) D. 8(t) - {ian (1) tan(t) t>3 B. u(t-3)(tan(t)-4) +4 C. 4u(t-3) + tan(t)u(t-3) D. 3te +et D. √2
Transcribed Image Text:A. x² = y² +1 3. A. 4. ( 5. A. y cie + c₂e6 3) The solution of = 5, y(1) = 1 is. B. y2x2+1 A. 8. A. y = ₁x + ₂x² us) The general solution of y" + 5y' -6y = 0 is. B. y cie-6 + c₂e 9. ( The integrating factor of xy' + 2y = is. C. ³ B. 2 A. e-t -) The general solution of r2y" - 3xy' + 4y = 0 is. B. y cr² + c₂r²ln(r) The radius of convergence of 1(-1)*3kk(x-4). B. 0 C. 00 11. ( A. 3tet + et D. 3 6. (s) The minimum radius of convergence of power series solution about r= -2 of the ODE (x²-49)y" + 2xy' + y = 0.. A. 7 B. 8 s) The value of {} is. B. et 844 3) The value of L-1{} is. B. 2tet + et A. 1 A. e²(+²) - C. y² = 2x² +1 The value of L{u(t-2)} is. B. e²(++) 10. (The equivalent form of Using unit step functions is A. u(t-3) (tan(t) + 4) + 4 D. 4u(t-3)-tan(t)u(t-3) C. ycie + c₂e-bx f(t) = C. y=+(2) C. 6 The value of f tan(t)o(t-)dt B. 0 C. u(t) C. 2te-t + et D. ²-x²=0. 0<t<3 D. D. y cez + cela C. ₂ D. y = 2+2 D. 5 C. e(++) D. e² (3+²) D. 8(t) - {ian (1) tan(t) t>3 B. u(t-3)(tan(t)-4) +4 C. 4u(t-3) + tan(t)u(t-3) D. 3te +et D. √2
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