Solve the initial value problem y" – 2y' + 5y = 0; y(0) = 2, y'(0)...
Solve 2y'' – 5y' – 25y = 0, y(0) = -6, y'(0) = – 15 (t) = Consider the initial value problem y' + 3y' – 10y = 0, y(0) = a, y'(0) = 3 Find the value of a so that the solution to the initial value problem approaches zero as t + oo a = 1
21. Solve the initial value problem y" - y-2y= 0, y(0) = a , y ( 0) the solution approaches zero as t 0o. 2. Then find a so that
Solve the initial value problem y" + 3y' + 2y = 8(t – 3), y(0) = 2, y'(0) = -2. Answer: y = u3(t) e-(-3) - u3(t)e-2(1-3) + 2e-, y(t) ={ 2e-, t<3, -e-24+6 +2e-l, t>3. 5. [18pt] b) Solve the initial value problem y' (t) = cost + Laplace transforms. +5° 867). cos (t – 7)ds, y(0) – 1 by means of Answer:
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
Problem 2. (a) Solve the initial value problem I y' + 2y = g(t), 1 y(0) = 0, where where | 1 if t < 1, g(t) = { 10 if t > 1 (t) = { for all t. Is this solution unique for all time? Is it unique for any time? Does this contradict the existence and uniqueness theorem? Explain. (b) If the initial condition y(0) = 0 were replaced with y(1) = 0, would there necessarily be...
differential equations .. Boundary Value. Solve the following: y" + 2y' - 5y = 0, y(0) = 0, y'(1) = 0 F. Boundary Value. Solve the following: y" + 2y' - 3y = 9x, y(0) = 1, y'(1) = 2
(4 points) Use the Laplace transform to solve the following initial value problem: y" – 2y + 5y = 0 y(0) = 0, y'(0) = 8 First, using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}| find the equation you get by taking the Laplace transform of the differential equation = 01 Now solve for Y(3) By completing the square in the denominator and inverting the transform, find g(t) =
Find the solution of the initial value problem y" – 2y' + 5y = g(t), y(0) = 0, y'(0) = 0, where g(t) is a continuous, otherwise arbitrary, function. Oy(t) = g(t) 1 y(t) = (sets sin(2t))g(t) Oy(t) = (cos(2t)) * g(t) Oy(t) = (cos(2t))g(t) y(t) = (1 e*) + f(t) x(t) =() e sin(26)g(t) g(t) = ( e sin(2t) + (t) y(t) = Ce+ sin(2t)) *g(t) 1
5.Solve the initial value problem y" +5y' +6y-g(t), y(0) 0,(0) 2, where (t)-t 1<t<5,. 1, 5 < t. Then sketch the graph of the solution. (Use technologies. Be sure the graph is neat.) Sec. 7.6.39]
#6 Solve the initial value problem y(0)- 2, y,(0) 1 y"-3y' + 2y-6(t-3);