Question

Let \( y(t) \) be the solution of the initial value problem \[ \frac{d^2 y}{dt^2} + a \frac{dy}{dt} + b y = f(t), a > 0, b > 0, a \neq b, a^2 - 4b = 0, \] with initial conditions \( y(0) = 0 \), \( \frac{dy}{dt}(0) = 0 \), obtained by the method of Laplace transform. Then

• A. \( y(t) = \int_0^t \tau e^{-\frac{a \tau}{2}} f(t - \tau) \, d\tau \)
• B. \( y(t) = \int_0^t e^{-\frac{a \tau}{2}} f(t - \tau) \, d\tau \)
• C. \( y(t) = \int_0^t \tau e^{-\frac{b \tau}{2}} f(t - \tau) \, d\tau \)
• D. \( y(t) = \int_0^t e^{-\frac{b \tau}{2}} f(t - \tau) \, d\tau \)

Hint:

For second-order linear differential equations with constant coefficients, the solution can be found using the Laplace transform, and particular solutions can be derived through integration.

Answer 1

The Correct Option is A

The given second-order linear differential equation is: \[ \frac{d^2 y}{dt^2} + a \frac{dy}{dt} + b y = f(t), \] with initial conditions \( y(0) = 0 \) and \( \frac{dy}{dt}(0) = 0 \). The characteristic equation for this second-order linear differential equation is: \[ r^2 + a r + b = 0. \] Given the condition \( a^2 - 4b = 0 \), the roots of the characteristic equation are: \[ r = -\frac{a}{2}. \] This means the general solution will have a form involving an exponential decay term \( e^{-\frac{a}{2}t} \), and the particular solution will be obtained using the Laplace transform. The solution can be written as: \[ y(t) = \int_0^t \tau e^{-\frac{a \tau}{2}} f(t - \tau) \, d\tau, \] which corresponds to option (A).