Find the solution of the IVP using Laplace Transform a) \frac{d^2y}{dt^2} + 6\frac{dy}{dt} + 13y = 13u_4(t), y(0) = 3, y'(0) = 1 b) \frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 9y = 20u_2(t)sin(t), y(0) = 1, y'(0) = 2
The Correct Answer and Explanation is:
We are asked to solve two initial value problems (IVPs) using Laplace Transforms.
Problem (a):
d2ydt2+6dydt+13y=13u4(t),y(0)=3,y′(0)=1\frac{d^2y}{dt^2} + 6\frac{dy}{dt} + 13y = 13u_4(t), \quad y(0) = 3, \quad y'(0) = 1dt2d2y+6dtdy+13y=13u4(t),y(0)=3,y′(0)=1
Step 1: Take Laplace Transforms of both sides
Let Y(s)Y(s)Y(s) be the Laplace Transform of y(t)y(t)y(t).
Use known transforms:L{y′′(t)}=s2Y(s)−sy(0)−y′(0)\mathcal{L}\left\{y”(t)\right\} = s^2Y(s) – sy(0) – y'(0)L{y′′(t)}=s2Y(s)−sy(0)−y′(0)L{y′(t)}=sY(s)−y(0)\mathcal{L}\left\{y'(t)\right\} = sY(s) – y(0)L{y′(t)}=sY(s)−y(0)L{u4(t)}=e−4ss\mathcal{L}\left\{u_4(t)\right\} = \frac{e^{-4s}}{s}L{u4(t)}=se−4s
Apply the Laplace Transform:s2Y(s)−3s−1+6(sY(s)−3)+13Y(s)=13⋅e−4sss^2Y(s) – 3s – 1 + 6(sY(s) – 3) + 13Y(s) = 13 \cdot \frac{e^{-4s}}{s}s2Y(s)−3s−1+6(sY(s)−3)+13Y(s)=13⋅se−4s
Simplify:s2Y(s)−3s−1+6sY(s)−18+13Y(s)=13e−4sss^2Y(s) – 3s – 1 + 6sY(s) – 18 + 13Y(s) = \frac{13e^{-4s}}{s}s2Y(s)−3s−1+6sY(s)−18+13Y(s)=s13e−4s
Combine like terms:(s2+6s+13)Y(s)=13e−4ss+3s+19(s^2 + 6s + 13)Y(s) = \frac{13e^{-4s}}{s} + 3s + 19(s2+6s+13)Y(s)=s13e−4s+3s+19
Solve for Y(s)Y(s)Y(s):Y(s)=13e−4ss(s2+6s+13)+3s+19s2+6s+13Y(s) = \frac{13e^{-4s}}{s(s^2 + 6s + 13)} + \frac{3s + 19}{s^2 + 6s + 13}Y(s)=s(s2+6s+13)13e−4s+s2+6s+133s+19
Now take the inverse Laplace of each term separately using known formulas and shift theorems.
Problem (b):
d2ydt2+4dydt+9y=20u2(t)sin(t),y(0)=1,y′(0)=2\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 9y = 20u_2(t)\sin(t), \quad y(0) = 1, \quad y'(0) = 2dt2d2y+4dtdy+9y=20u2(t)sin(t),y(0)=1,y′(0)=2
Step 1: Take Laplace TransformsL{y′′(t)}+4L{y′(t)}+9L{y(t)}=20L{u2(t)sin(t)}\mathcal{L}\{y”(t)\} + 4\mathcal{L}\{y'(t)\} + 9\mathcal{L}\{y(t)\} = 20\mathcal{L}\{u_2(t)\sin(t)\}L{y′′(t)}+4L{y′(t)}+9L{y(t)}=20L{u2(t)sin(t)}
Use:L{sin(t)u2(t)}=e−2s⋅1s2+1\mathcal{L}\{\sin(t)u_2(t)\} = e^{-2s} \cdot \frac{1}{s^2 + 1}L{sin(t)u2(t)}=e−2s⋅s2+11
So:20⋅e−2ss2+120 \cdot \frac{e^{-2s}}{s^2 + 1}20⋅s2+1e−2s
Now, apply Laplace to LHS:s2Y(s)−sy(0)−y′(0)+4(sY(s)−y(0))+9Y(s)=20e−2ss2+1s^2Y(s) – sy(0) – y'(0) + 4(sY(s) – y(0)) + 9Y(s) = \frac{20e^{-2s}}{s^2 + 1}s2Y(s)−sy(0)−y′(0)+4(sY(s)−y(0))+9Y(s)=s2+120e−2s
Substitute initial values:s2Y(s)−s−2+4sY(s)−4+9Y(s)=20e−2ss2+1s^2Y(s) – s – 2 + 4sY(s) – 4 + 9Y(s) = \frac{20e^{-2s}}{s^2 + 1}s2Y(s)−s−2+4sY(s)−4+9Y(s)=s2+120e−2s
Simplify:(s2+4s+9)Y(s)=20e−2ss2+1+s+6(s^2 + 4s + 9)Y(s) = \frac{20e^{-2s}}{s^2 + 1} + s + 6(s2+4s+9)Y(s)=s2+120e−2s+s+6
Solve for Y(s)Y(s)Y(s):Y(s)=20e−2s(s2+1)(s2+4s+9)+s+6s2+4s+9Y(s) = \frac{20e^{-2s}}{(s^2 + 1)(s^2 + 4s + 9)} + \frac{s + 6}{s^2 + 4s + 9}Y(s)=(s2+1)(s2+4s+9)20e−2s+s2+4s+9s+6
Take inverse Laplace term by term using partial fraction decomposition and known transforms. The shifting property applies due to the e−2se^{-2s}e−2s factor, which shifts the function by 2 units in time.
These solutions represent the response of the systems to a unit step or shifted sine input, filtered through a second-order differential equation. Laplace Transform simplifies solving such linear time-invariant differential equations by converting them into algebraic equations in the frequency domain.
