Let y(t) be the solution of the initial value problem (d ^ 2 * y)/(d * t ^ 2) + a * d/dt (y) + by = f(t) a > 0 y(0) = 0 b > 0 d/dt (y) * (0) = 0 obtained by the method of Laplace transform. Then a ne b, a ^ 2 - 4b = 0(A) y(t) = integrate tau * e ^ ((- a * tau)/2) * f(t - tau) dtau from 0 to t(B) y(t) = integrate e ^ ((- a * tau)/2) * f(t - tau) dtau from 0 to t(C)y(t) = integrate tau * e ^ ((- b * tau)/2) * f(t - tau) dtau from 0 to t(D)y(t) = integrate e ^ ((- b * tau)/2) * f(t - tau) dtau from 0 to t

The given differential equation is a second order linear homogeneous differential equation with constant coefficients. The Laplace transform is a powerful method for solving such equations.

The Laplace transform of the given differential equation is:

L{y''(t) + ay'(t) + by(t)} = L{f(t)}

Using the properties of the Laplace transform, this becomes:

s^2Y(s) - sy(0) - y'(0) + a[sY(s) - y(0)] + bY(s) = F(s)

Given that y(0) = 0 and y'(0) = 0, the equation simplifies to:

(s^2 + as + b)Y(s) = F(s)

Solving for Y(s), we get:

Y(s) = F(s) / (s^2 + as + b)

Taking the inverse Laplace transform of both sides gives the solution y(t) in the time domain.

Given the conditions a ≠ b and a^2 - 4b = 0, the roots of the characteristic equation s^2 + as + b = 0 are complex and equal. Therefore, the solution y(t) will be of the form:

y(t) = e^(-at/2) * ∫(e^(at/2) * f(t - τ) dτ) from 0 to t

This corresponds to option (B) in the given choices.