Use the Laplace transform to solve the given initial-value problem. y" + 6y' + 5y = 0, y(0) = 1, y'(O) = 0 y(t) =
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]
3(10pt). Solve the initial value problem: y(3) – 5y" + 6y' = 0, y(0) = 1, y(0) = 0, y" (0) = 1.
SOLVE THE FOLLOWING SYSTEM OF EQUATIONS BY THE CRAMER'S METHOD 3X+5Y+3Z-12 2X+5Y-2Z-6 3x+6Y+3Z-3 a) X Y b) CHECK YOUR RESULTS. (USE MATRICE FUNCTIONS, PRESS F2. AND THEN PRESS CTRL+SHIFT+ENTER) 3IF Y-SINC) EXPOO. INTEGRATE Y FROM X-0 Tox-1. COMPARE WITH REAL VALUE IF DX-0 a) INT b) INT ,IF DX- 005 REAL VALUE 3) Plot sin x letting maco c/ Prepave hese cuves 4) SOLVE THE FOLLOWING SYSTEM OF EQUATIONS BY INVERSE METHOD 3 X+3Z-13 2X +5 Y-2Z-2 3 X+6Y+2Z-3 Z-...
13. Use the Laplace transform to solve the initial value problem: (&pts) y" - 6y' + 5y = 3e, y(0) = 2, 7(0) = 3
Q1) Solve the following DE: (Using Laplace transform is recommended) y" + 5y' – 6y = f(t), y(0) = 0, y'(0) = 0, where 0 <t< 2 f(t) = {-4 t>2 1
(1 point) Use the Laplace transform to solve the following initial value problem: y" + 6y' - 16y = 0 y(0) = 3, y(0) = 1 First, using Y for the Laplace transform of y(t), i.e., Y = C{y(t)). find the equation you get by taking the Laplace transform of the differential equation = 0 Now solve for Y(s) = and write the above answer in its partial fraction decomposition, Y(S) = Y(s) = A. where a <b Now by...
Solve the following differential equation with given initial conditions using the Laplace transform. y" + 5y' + 6y = ut - 1) - 5(t - 2) with y(0) -2 and y'(0) = 5. 1 AB I
3. Solve the recursion equation: y[n] – 5y[n – 1] + 6y[n – 2] = 2 U[n] with y[-1] = 6 and y[-2] = 4
Tutorial Exercise Use the Laplace transform to solve the given initial-value problem. y' + 5y = et (0) = 2 Step 1 To use the Laplace transform to solve the given initial value problem, we first take the transform of each member of the differential equation + 6y et The strategy is that the new equation can be solved for ty) algebraically. Once solved, transforming back to an equation for gives the solution we need to the original differential equation....