The Correct Answer and Explanation is:
The given initial value problem is:
y′+y=e−t,y(0)=1y’ + y = e^{-t}, \quad y(0) = 1
To solve this using the Laplace transform, we follow these steps.
Step 1: Apply the Laplace Transform
Taking the Laplace transform of both sides,
L[y′]+L[y]=L[e−t]\mathcal{L}[y’] + \mathcal{L}[y] = \mathcal{L}[e^{-t}]
Using properties of Laplace transforms:
sY(s)−y(0)+Y(s)=1s+1sY(s) – y(0) + Y(s) = \frac{1}{s+1}
Substituting y(0)=1y(0) = 1,
sY(s)−1+Y(s)=1s+1sY(s) – 1 + Y(s) = \frac{1}{s+1}
Step 2: Solve for Y(s)Y(s)
Rearrange the equation:
(s+1)Y(s)=1s+1+1(s+1)Y(s) = \frac{1}{s+1} + 1
Y(s)=1(s+1)2+1s+1Y(s) = \frac{1}{(s+1)^2} + \frac{1}{s+1}
Step 3: Take the Inverse Laplace Transform
Using inverse Laplace transforms:
L−1[1(s+1)2]=te−t\mathcal{L}^{-1} \left[\frac{1}{(s+1)^2}\right] = t e^{-t}
L−1[1s+1]=e−t\mathcal{L}^{-1} \left[\frac{1}{s+1}\right] = e^{-t}
Adding both results:
y(t)=te−t+e−ty(t) = t e^{-t} + e^{-t}
Explanation
Laplace transforms simplify solving differential equations by converting them into algebraic equations. Applying the Laplace transform to the differential equation converts differentiation into multiplication by ss. Using properties of transforms, we rewrite the equation in terms of Y(s)Y(s), which represents the transformed function. After algebraic manipulation, we isolate Y(s)Y(s) and determine its inverse transform to find y(t)y(t). The result y(t)=e−t(t+1)y(t) = e^{-t} (t+1) satisfies the given differential equation and initial condition.
