Question ; Convert following second order ode into a system of first order ODEs
\[\frac{d^2y}{dt^2}+\frac{dy}{dt}+y=2t+3\sin{\left(t\right)} , y\left(0\right)=2,\ y^\prime\left(0\right)=3 \]
Solution :
Given second order ODE \[\frac{d^2y}{dt^2}+\frac{dy}{dt}+y=2t+3\sin{\left(t\right)} \] (1) Subject to intial conditions \[\ \ y\left(0\right)=2,\ y^\prime\left(0\right)=3\] To convert it into a system of first order ODEs Substitute \[y_1=y,\ y_2=\frac{dy}{dt}\] Differentiating y1 , we get => \[\frac{dy_1}{dt}=\frac{dy}{dt}=y_2=f_1\left(t,\ y_1,y_2\right)\] And Differentiating y2 , we get \[\frac{dy_2}{dt}=\frac{d^2y}{dt^2}\] We shall use eqn (1), to get following \[\frac{d^2y}{dt^2}=-\frac{dy}{dt}-y+2t+3\sin{\left(t\right)}=-y_2-y_1+3\sin{\left(t\right)}\] Thus system of first order odes is \[\frac{dy_1}{dt}=y_2=f_1\left(t,\ y_1,y_2\right)\] \[\frac{dy_2}{dt}=-y_2-y_1+3\sin{\left(t\right)}=f_2\left(t,\ y_1,y_2\right)\] Subject to initial conditions \[ y_1\left(0\right)=y\left(0\right)=2,\ y_2\left(0\right)=y^\prime\left(0\right)=3\]