Find the solution.y' = 5y + e 2x y2, y(0) = 2-2х139е 15 — Зе- 3e -2x \1/3Ay =1715х19е15x _ -3x \1/35eВy22739e Sx-2х \1/3— еC2513e-15x + 3e 2x )1/3'+3Dy =23

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Answered: Find the solution. y' = 5y + e 2x y2,…

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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Match column B with the correct answer on column A, write only the capital letter of the answer on the blank provided at the left side/right side of the test paper. Match from column A the solution of the differential equations on column B. Column A A. xlnx + ylny = c B. 2y = x4 + cx2 C. e(x^2) + y2 = c D. x2 = cy3 E. x2 y = 2x4 + c F. ylnx + xlny = c G. x2 (x2 + 2y2) = c H. x (y + x2) = c (y – 2x) I. xy2 – x2y + 3x2 – 2y = c J. xy3 = 2y2 + 4y + 4 + cey K. y (x2 + c) + 2 = 0 L. (y – x) (y + 3x)3 = cx3 M. x2 y = 2x2 + c N. x (y + 1) = (1 + cx) ey O. e(-x^2) + y-2 = c P. (y + x2) = c (y – 2x) Q. x2y2 + 2xy – x2 = c R. x4 = c2 (4x2 + y2) S. x2 (x+ 2y) = c T. x2 + y2 – 3xy = c Column B (x – y) (4x + y) dx + x (5x – y) dx = 0 xy3 dx + e(x^2) dy = 0 y dx + (3x – xy + 2) dy = 0 2 (2x2 + y2) dx – xy dy = 0 (x2 + y2) – xy dy = 0 (2x – 3y) dx + (2x – 3x) dy = 0 x2 yy’ = ey (x4 + 2y) dx – x dy = 0 (3x2 – 2xy + 3y2) dx = 4xydy
Match column B with the correct answer on column A, write only the capital letter of the answer on the blank provided at the left side/right side of the test paper. Match from column A the solution of the differential equations on column B. Column A A. xlnx + ylny = c B. 2y = x4 + cx2 C. e(x^2) + y2 = c D. x2 = cy3 E. x2 y = 2x4 + c F. ylnx + xlny = c G. x2 (x2 + 2y2) = c H. x (y + x2) = c (y – 2x) I. xy2 – x2y + 3x2 – 2y = c J. xy3 = 2y2 + 4y + 4 + cey K. y (x2 + c) + 2 = 0 L. (y – x) (y + 3x)3 = cx3 M. x2 y = 2x2 + c N. x (y + 1) = (1 + cx) ey O. e(-x^2) + y-2 = c P. (y + x2) = c (y – 2x) Q. x2y2 + 2xy – x2 = c R. x4 = c2 (4x2 + y2) S. x2 (x+ 2y) = c T. x2 + y2 – 3xy = c Column B (x – y) (4x + y) dx + x (5x – y) dx = 0 xy3 dx + e(x^2) dy = 0 y dx + (3x – xy + 2) dy = 0 2 (2x2 + y2) dx – xy dy = 0 (x2 + y2) – xy dy = 0 (2x – 3y) dx + (2x – 3x) dy = 0 x2 yy’ = ey (x4 + 2y) dx – x dy = 0 (3x2 – 2xy + 3y2) dx = 4xydy (1 + lnx) dx + (1 + lny) dy = 0
Match column B with the correct answer on column A, write only the capital letter of the answer on the blank provided at the left side/right side of the test paper. Match from column A the solution of the differential equations on column B. Column A A. xlnx + ylny = c B. 2y = x4 + cx2 C. e(x^2) + y2 = c D. x2 = cy3 E. x2 y = 2x4 + c F. ylnx + xlny = c G. x2 (x2 + 2y2) = c H. x (y + x2) = c (y – 2x) I. xy2 – x2y + 3x2 – 2y = c J. xy3 = 2y2 + 4y + 4 + cey K. y (x2 + c) + 2 = 0 L. (y – x) (y + 3x)3 = cx3 M. x2 y = 2x2 + c N. x (y + 1) = (1 + cx) ey O. e(-x^2) + y-2 = c P. (y + x2) = c (y – 2x) Q. x2y2 + 2xy – x2 = c R. x4 = c2 (4x2 + y2) S. x2 (x+ 2y) = c T. x2 + y2 – 3xy = c Column B 2y dx = 3x dy (y2 – 2xy + 6x) dx – (x2 – 2xy + 2) dy = 0 2(y – 4x2) dx + xdy = 0 y' = xy2 (xy2 + y – x) dx + x (xy + 1) dy = 0 (x – y) (4x + y) dx + x (5x – y) dx = 0 xy3 dx + e(x^2) dy = 0 y dx + (3x – xy + 2) dy = 0 2 (2x2 + y2) dx – xy dy = 0 (x2 + y2) – xy dy = 0 (2x – 3y) dx + (2x – 3x) dy = 0 x2 yy’ =…
Find a homogeneous linear differential equation with constant coefficients whose general solution is given: y = C1cosh 7x + C2sinh7x y = C1 + C2e2xcos 5x + C3e2xsin 5x y = C1cos x + C2sin x + C3cos 2x + C4sin 2x
solve the for the general solution of the differential equation( linear homologous equation):(4D4 - 4D3 - 23D2 + 12D +36 )y =0
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solve the for the general solution of the differential equation( linear homologous equation):y''- 9y= 0 with x=0, y=0 and y'=3
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Suppose thay a drug is administered to a person in a single dose, and assume that the drug does not accumulate in body tissue, but is excerted through urine. Denote the amount of drug in the body at time t by x1(t) and in the urine at time t by x2(t). If x1(0)=6 mg and x2(0)=0, can you find a system of differential equations for x1(t) and x2(t) if it takes 20 minutes for the drug to be at one-half of its initial amount in the body?
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Find the solution. y' = 5y + e 2x y2, y(0) = 2 -2х 139е 15 — Зе - 3e -2x \1/3 A y = 17 15х 19е 15x _ -3x \1/3 5e В y2 27 39e Sx -2х \1/3 — е C 25 13e-15x + 3e 2x )1/3 '+3 D y = 23