Question

Consider the differential equation $$ \frac{dy}{dx} + a y = e^{-bt} \quad \text{with} \quad y(0) = 0. $$ Then the Laplace transform of $ y(t) $ is

• A. \( \frac{1}{(s+a)(s+b)} \)
• B. \( \frac{1}{(s+a)b} \)
• C. \( \frac{1}{a(s+b)} \)
• D. \( \frac{e^{-a} - e^{-b}}{b - a} \)

Hint:

To solve a first-order linear differential equation using the Laplace transform, take the transform of both sides and solve for the Laplace of the unknown function.

Answer 1

The Correct Option is A

The given differential equation is: \[ \frac{dy}{dx} + a y = e^{-bt} \] We can take the Laplace transform of both sides. For the left-hand side, we apply the standard Laplace transform of the derivative and the function: \[ \mathcal{L} \left\{ \frac{dy}{dx} \right\} = s Y(s) - y(0) \quad \text{(where \( y(0) = 0 \))} \] Thus, the Laplace transform of the left-hand side becomes: \[ s Y(s) + a Y(s) = (s + a) Y(s) \] For the right-hand side, the Laplace transform of \( e^{-bt} \) is: \[ \mathcal{L} \left\{ e^{-bt} \right\} = \frac{1}{s + b} \] Therefore, the Laplace-transformed equation becomes: \[ (s + a) Y(s) = \frac{1}{s + b} \] Solving for \( Y(s) \): \[ Y(s) = \frac{1}{(s + a)(s + b)} \]
Conclusion: The Laplace transform of \( y(t) \) is \( \frac{1}{(s+a)(s+b)} \).