So, I was given $ \frac{dx}{dt}=4cos(t) $ and $ \frac{dy}{dt}=sin(t) $ at $ 0\le t \le \frac{π}{2}$
I was told that the particle is at the origin at t = 0, and to find x(t) and y(t), the position of the particle.
Am I just supposed to do $\int4cos(t)$ and $\int sin(t)$, which equals to $4sin(t)+C$ and $-cos(t)+C$, or is there another method?
Given simultaneous ordinary differential equations are:
$$ {dx \over dt} = 4 \cos t, \ \ {dy \over dt} = \sin t \tag{1}$$
which are defined over the time interval $\left[ 0, {\pi \over 2} \right].$
Integrating (1) with respect to $t$, we get $$ x(t) = 4 \sin t + c_1, \ \ y(t) = - \cos t + c_2 \tag{2} $$ where $c_1, c_2$ are arbitrary constants of integration.
To find $c_1, c_2$, we use the information given in the question, namely, the particle is at the origin at $t = 0$.
This means: When $t = 0$, $x(t) = 0$ and $y(t) = 0$.
Substituting $x(0) = 0$ and $y(0) = 0$ in (2), we get $$ c_1 = 0, \ \ c_2 = 1 \tag{3} $$
Substituting the values of $c_1$ and $c_2$ from (3) into (2), we get the initial solution of the simultaneous ODEs (1) as $$ x(t) = 4 \sin t, \ \ y(t) = 1 - \cos t, \ \ \ \ \mbox{where} \ \ t \in \left[ 0, {\pi \over 2} \right]. \tag{4} $$