Answer to Question #212974 in Differential Equations for eve

Question

given the system express y as a function of t

1. 3dx/dt+2dy/dt-x+y=t-1
2. dx/dt+dy/dt-x=t+2

Answer 1

Given the system let us express $y$ as a function of $t.$ Let us add to $3\frac{dx}{dt}+2\frac{dy}{dt}-x+y=t-1$ the equation $\frac{dx}{dt}+\frac{dy}{dt}-x=t+2$ multiplied by $-3.$ We have that $-\frac{dy}{dt}+2x+y=-2t-7.$ It follows that $x=\frac{1}{2}\frac{dy}{dt}-\frac{1}{2}y-t-\frac{7}{2},$ and hence $\frac{dx}{dt}=\frac{1}{2}\frac{d^2y}{dt^2}-\frac{1}{2}\frac{dy}{dt}-1.$ We get the equation $\frac{1}{2}\frac{d^2y}{dt^2}-\frac{1}{2}\frac{dy}{dt}-1+\frac{dy}{dt}-(\frac{1}{2}\frac{dy}{dt}-\frac{1}{2}y-t-\frac{7}{2})=t+2$ which is equivalent to $\frac{1}{2}\frac{d^2y}{dt^2}+\frac{1}{2}y=-\frac{1}{2},$ and hence to $\frac{d^2y}{dt^2}+y=-1.$ The characteristic equation $k^2+1=0$ have the roots $k_1=i$ and $k_2=-i.$ It follows that the solution is of the form $y(t)=C_1\cos t+C_2\sin t+y_p,$ where $y_p=a.$ It follows that $\frac{d^2y_p}{dt^2}=0,$ and hence $a=-1.$ Therefore, the solution is the following:

$y(t)=C_1\cos t+C_2\sin t-1.$