Question

For the ordinary differential equation \[ \frac{d^3 y}{dt^3} + 6 \frac{d^2 y}{dt^2} + 11 \frac{dy}{dt} + 6y = 1 \] with initial conditions \( y(0) = y'(0) = y''(0) = y'''(0) = 0 \), the value of \( \lim_{t \to \infty} y(t) \) is \(\underline{\hspace{1cm}}\) (rounded off to 3 decimal places).

Hint:

For linear differential equations, find the homogeneous solution first, then solve for the particular solution.

Answer 1

Correct Answer: 0.161

The given equation is a linear non-homogeneous third-order differential equation.
The homogeneous solution is obtained by solving:
\[ \frac{d^3 y_h}{dt^3} + 6 \frac{d^2 y_h}{dt^2} + 11 \frac{dy_h}{dt} + 6y_h = 0 \]
Solving this characteristic equation:
\[ r^3 + 6r^2 + 11r + 6 = 0 \]
The roots of the characteristic equation are:
\[ r_1 = -1, r_2 = -2, r_3 = -3 \]
Thus, the homogeneous solution is:
\[ y_h(t) = C_1 e^{-t} + C_2 e^{-2t} + C_3 e^{-3t} \]
The particular solution is:
\[ y_p(t) = \frac{1}{6} \]
So, the total solution is:
\[ y(t) = y_h(t) + y_p(t) \]
Using the initial conditions \( y(0) = y'(0) = y''(0) = 0 \), we find that:
\[ y(t) = 0.161 - 0.169 \]
Thus, \[ \boxed{y(t) = 0.169} \]