So 0<t<5 Using the Laplace transform, solve the initial value problem y' + y = 3...
Use the Laplace transform to solve the given initial-value problem. so, 0 <t< 1 y' + y = f(t), y(0) = 0, where f(t) 17, t21 y(t) = + ult-
(1 point) Consider the initial value problem y' + 3y = 0 if 0 <t <3 9 if 3 < t < 5 0 if 5 <t< oo, y(0) = 3. (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(s)(5+6)...
2. Use the Laplace Transform to solve the initial value problem y"-3y'+2y=h(t), y(O)=0, y'(0)=0, where h (t) = { 0,0<t<4 2, t>4
QUESTION 3 Use Laplace Transform to solve the initial value problem y" + 9y = f(t) ,y(0) = 1, y'(0) = 3 where 6, f(t) 0 <t<nt i < t < 0
Find the Laplace transform Y(s) = L{y} of the solution of the given initial value problem: 1, y' + 9 = 0<t<T 0,7 <t< y(0) = 5, y'(0) = 4
4. Use the Laplace transform to solve the initial value problem y" + y = f(1) = -2, ost<2 13t+4, 122 y(0) = 0, y'(0) = -1
15) 5. Use Laplace transforms to solve the initial value problem y" + y = g(t), y'(0) = 0, y(0) = 0, where 0 St< 10, 10 t 20, 0, g(t) = (t-10), 1, t < 20, and describe the qualitative behavior of the solution fort 20
(1 point) Take the Laplace transform of the following initial value problem and solve for Y(8) = L{y(t)}; ſ1, 0<t<1 y" – 6y' - 27y= { O, 1<t y(0) = 0, y'(0) = 0 Y(8) = (1-e^(-s)(s(s^2-6s-27)) Now find the inverse transform: y(t) = (Notation: write uſt-c) for the Heaviside step function uct) with step at t = c.) Note: 1 | 1 s(8 – 9)(8 + 3) 36 6 10 + s $+37108 8-9
Find the Laplace transform of the given function Solve the integral equation f(t) = { 0 < t < 2 t 22 t y(t) = 4t – 3 y(z)sin(t – z)dz 0
cometeness and clarity please ! 1. Solve the initial value problem using Laplace transforms. ſi ost<5 y" - 5y + 4y = 0 t25 y(0) = 0, 7(0) = 1