Problem #6: Consider the following initial value problem.0 t ≤ 390у' + бу =3 ≤ t < 77 ≤ t < 00y(0) = 2(a) Find the Laplace transform of the right hand side of the above differential equation.(b) Let y(t) denote the solution to the above differential equation, and let Y((s) denote the Laplace transform ofy(t). Find Y(s).(c) By taking the inverse Laplace transform of your answer to (b), the solution y(t) can be written in the formf(t) U(t − 3) + g (t) U (t − 7) + h(t). Enter the function f(t) into the answer box below.(d) Referring to part (c) above, enter the function g(t) into the answer box below.(e) Referring to part (c) above, enter the function h(t) into the answer box below.

Introduction: At first glance, the Laplace Transform appears to be quite an arcane and abstract…

Answered: Problem #6: Consider the following…

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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Consider the initial value problem if 0 < t < 1 y + 3y = if 1 < t < 5 if 5
1. Find the general solution of the differential equation y" - y = e²t using (a) a method (without Laplace transform) given in Chapter 3 or 4; (b) the Laplace transform. Hint: In order to use the Laplace transform, you need to use the initial conditions as y(0) = a and y'(0) = b, where a, b are real (arbitrary) constants. 2. Find the solution of the initial value problem y" + 4y = g(t) = { 1° 1 > 16 0. y" + by' + cy= gi(t); y(0) = a, y'(0) = ß on interval 0 to obtain the solution, say, y2(t). (You are able to evaluate y₁ (4) and y(), the initial values at t = 1, since you have obtained y₁ (t) already.) The solution of the original IVP then is y(t) = { vh(t), 0
Consider the initial value problem y' + 3y = (b) Solve your equation for Y. Y = L {y} 0 11 0 (a) Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y by Y. Do not move any terms from one side of the equation to the other (until you get to part (b) below). = y = if 0 < t < 1 if 1 < t < 6 if 6 < t <∞, y(0) = 10. = (c) Take the inverse Laplace transform of both sides of the previous equation to solve for y.
Consider the initial value problem where g(t) = y" + 36y = g(t), if 0 < t < 5 if 5 < t < ∞. a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t) by Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) below). (s^2+36)Y(s) 1/s^2 b. Solve your equation for Y(s). Y(s) = L{y(t)} 1/36[1/s^2-1/(s^2+36)] y(0) = 0, y'(0) = 0, = = help (formulas) c. Take the inverse Laplace transform of both sides of the previous equation to solve for y(t). y(t) = 1/36[t-sin(6t)/6]
Question 4. Let y(t) satisfy the following 2nd order ordinary differential equation: 3y" — y' - 4y = 1, with initial conditions: y(0) = −1, y'(0) = 9. Let Y(s) represent the Laplace Transform of y(t). Then Y(s) can be represented as: d (3s² +bs + c)Y(s) = +e+fs, S where b, c, d, e and f are constants. The above equation for Y(s) may be rearranged to give: ps² + qs + r Y(s) s(3s² + bs + c) where p, q and r are constants. = Enter b: Enter c: Enter d: Enter e: Enter f: Enter p: Enter q: Enter r: I
Solve using Methods of solution to First Order/High Order Differential EquationOr using Laplace Transform
Q8
Consider the initial value problem y" + 25y = cos(5t), y(0) = 6, y'(0) = 2. a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t) by Y (s). Do not move any terms from one side of the equation to the other (until you get to part (b) below). (formulas) b. Solve your equation for Y(s). Y(s) = L{y(t)} = = = y(t) = = help c. Take the inverse Laplace transform of both sides of the previous equation to solve for y(t).
I need help with this problem described below and an explanation for the solution. (Differential Equations)

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