Question

Consider ordinary differential equations given by \[ \frac{dx_1(t)}{dt} = 2x_2(t), \quad \frac{dx_2(t)}{dt} = r(t) \] with initial conditions \( x_1(0) = 1 \) and \( x_2(0) = 0 \). If \[ r(t) = \begin{cases} 1, & t \geq 0 \\ 0, & t < 0 \end{cases} \] then at \( t = 1 \), \( x_1(t) = \underline{2cm} \) (round off to the nearest integer).

Hint:

To solve a system of first-order ODEs with one depending on the other, solve the simpler equation first, then substitute into the next. Use the given initial conditions to evaluate the constants after integration.

Answer 1

We are given a system of ODEs: \[ \frac{dx_2(t)}{dt} = r(t) = 1 \quad {(for } t \geq 0{)} \] Integrate to find \( x_2(t) \): \[ x_2(t) = \int_0^t r(\tau) \, d\tau = \int_0^t 1 \, d\tau = t \] Now use \( x_2(t) = t \) in the first equation: \[ \frac{dx_1(t)}{dt} = 2x_2(t) = 2t \Rightarrow x_1(t) = \int_0^t 2\tau \, d\tau + x_1(0) = t^2 + 1 \] At \( t = 1 \): \[ x_1(1) = 1^2 + 1 = 2 \] \[ \boxed{x_1(1) = 2} \]