Use the Laplace transform to solve the given initial-value problem. y” + 4y = \delta (t – 2), y(0) = 1, y'(0) = 0 y(t) = \cos(2t) + U(t – 2) \cdot \frac{1}{2}\sin[2(t – 2)] y(t) = \frac{1}{2}U(t – 2)\sin(2t) y(t) = \frac{1}{2}U(t – 2)\sin[2(t – 2)] y(t) = \cos(2t) + U(t – 2) \cdot \frac{1}{2}\sin(2t) 5 pts
The Correct Answer and Explanation is:
To solve this initial-value problem using the Laplace transform, we need to take the Laplace transform of the differential equation and apply the initial conditions. Let’s break it down step by step.
The given equation is:
y′′+4y=δ(t−2)y” + 4y = \delta(t – 2)y′′+4y=δ(t−2)
with initial conditions:y(0)=1,y′(0)=0y(0) = 1, \quad y'(0) = 0y(0)=1,y′(0)=0
where δ(t−2)\delta(t – 2)δ(t−2) is the Dirac delta function, which represents an impulse at t=2t = 2t=2.
Step 1: Apply the Laplace Transform to the differential equation
The Laplace transform of the second derivative y′′(t)y”(t)y′′(t) is given by:L{y′′(t)}=s2Y(s)−sy(0)−y′(0)\mathcal{L}\{y”(t)\} = s^2 Y(s) – sy(0) – y'(0)L{y′′(t)}=s2Y(s)−sy(0)−y′(0)
The Laplace transform of y(t)y(t)y(t) is Y(s)Y(s)Y(s), and the Laplace transform of the Dirac delta function δ(t−2)\delta(t – 2)δ(t−2) is:L{δ(t−2)}=e−2s\mathcal{L}\{\delta(t – 2)\} = e^{-2s}L{δ(t−2)}=e−2s
Substituting into the equation:(s2Y(s)−s⋅1−0)+4Y(s)=e−2s(s^2 Y(s) – s \cdot 1 – 0) + 4Y(s) = e^{-2s}(s2Y(s)−s⋅1−0)+4Y(s)=e−2s
Simplifying:(s2+4)Y(s)−s=e−2s(s^2 + 4)Y(s) – s = e^{-2s}(s2+4)Y(s)−s=e−2s
Rearranging for Y(s)Y(s)Y(s):Y(s)=s+e−2ss2+4Y(s) = \frac{s + e^{-2s}}{s^2 + 4}Y(s)=s2+4s+e−2s
Step 2: Inverse Laplace Transform
Now, we need to take the inverse Laplace transform of Y(s)Y(s)Y(s). This expression can be split into two parts:Y(s)=ss2+4+e−2ss2+4Y(s) = \frac{s}{s^2 + 4} + \frac{e^{-2s}}{s^2 + 4}Y(s)=s2+4s+s2+4e−2s
- Inverse Laplace Transform of ss2+4\frac{s}{s^2 + 4}s2+4s: L−1{ss2+4}=cos(2t)\mathcal{L}^{-1}\left\{\frac{s}{s^2 + 4}\right\} = \cos(2t)L−1{s2+4s}=cos(2t)
- Inverse Laplace Transform of e−2ss2+4\frac{e^{-2s}}{s^2 + 4}s2+4e−2s:
This is the Laplace transform of a shifted function: L−1{e−2ss2+4}=U(t−2)⋅sin[2(t−2)]\mathcal{L}^{-1}\left\{\frac{e^{-2s}}{s^2 + 4}\right\} = U(t – 2) \cdot \sin[2(t – 2)]L−1{s2+4e−2s}=U(t−2)⋅sin[2(t−2)] where U(t−2)U(t – 2)U(t−2) is the Heaviside step function.
Step 3: Combine the Results
Combining these results, we have:y(t)=cos(2t)+U(t−2)⋅sin[2(t−2)]y(t) = \cos(2t) + U(t – 2) \cdot \sin[2(t – 2)]y(t)=cos(2t)+U(t−2)⋅sin[2(t−2)]
This is the solution to the given initial-value problem. The term U(t−2)U(t – 2)U(t−2) shifts the sine wave to start at t=2t = 2t=2.
Final Answer:
y(t)=cos(2t)+U(t−2)⋅sin[2(t−2)]y(t) = \cos(2t) + U(t – 2) \cdot \sin[2(t – 2)]y(t)=cos(2t)+U(t−2)⋅sin[2(t−2)]
This solution is correct because the cosine term satisfies the equation before the impulse at t=2t = 2t=2, and the shifted sine term handles the effect of the delta function at t=2t = 2t=2.
