Take the Laplace transform of the following initial value problem and solve for Y(s)=L{y(t)}:
y′′−5y′−24y=S(t) y(0)=0,y′(0)=0
Where S(t)={1, ,0≤t<1 0, 1≤t<2} S(t+2)=S(t)
Y(s) = ?
first we find laplace transform of s(t) using heavy side function. ua(t) is the heavy side function with step size a.
L{ua(t)}=e-as/s
Take the Laplace transform of the following initial value problem and solve for Y(s)=L{y(t)}: y′′−5y′−24y=S(t) y(0)=0,y′(0)=0...
(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
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....
Use the Laplace transform to solve the given initial-value problem.y'' − 5y' = 8e4t − 4e−t, y(0) = 1, y'(0) = −1y(t) =
Use the Laplace transform to solve the given initial-value problem.y'' − 5y' = 8e4t − 4e−t, y(0) = 1, y'(0) = −1y(t) =
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y' +5y =12 - 8, 7(0) = 0, y'(0) = -2 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. Y(s)=0
Use the Laplace transform to solve the given initial-value problem. y" + 6y' + 5y = 0, y(0) = 1, y'(O) = 0 y(t) =
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y'' + 5y = 62 - 9, y(0) = 0, y'(0) = - 8 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. Y(s) =
where h is the Use the Laplace transform to solve the following initial value problem: y"+y + 2y = h(t – 5), y(0) = 2, y(0) = -1, Heaviside function. In the following parts, use h(t – c) for the shifted Heaviside function he(t) when necessary. a. First, take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation and then solve for L{y(t)}. L{y(t)}(s) = b. Express the solution y(t) as the...
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y' + 5y = 5t? -9, y(0) = 0, y'(0) = -3 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. 169=122 3.sin (1960) - cos (15) -
(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) =