Question

A system described by a linear, constant coefficient, ordinary, first order differential equation has an exact solution given by $y\left(t\right)$ for $t>0$, when the forcing function is $x\left(t\right)$ and the initial condition is $y\left(0\right)$. If one wishes to modify the system so that the solution becomes $-2y\left(t\right)$ for $t>0$, we need to

• A. change the initial condition to $-y\left(0\right)$ and the forcing funtion to $2x\left(t\right)$
• B. change the initial condition to $2y\left(0\right)$ and the forcing funtion to $-x\left(t\right)$
• C. change the initial condition to $j\sqrt{2}y\left(0\right)$ and the forcing function to $j\sqrt{2}x\left(t\right)$
• D. change the initial condition to $-2y\left(0\right)$ and the forcing function to $-2x\left(t\right)$

Answer 1

The correct option is D change the initial condition to $-2y\left(0\right)$ and the forcing function to $-2x\left(t\right)$

$\frac{dy\left(t\right)}{dt}+ky\left(t\right)=x\left(t\right)$
Taking Laplace transform of both sides, we have
$sY\left(s\right)-y\left(0\right)+kY\left(s\right)=X\left(s\right)$
$Y\left(s\right)[s+k]=X\left(s\right)+y\left(0\right)$
$\Rightarrow Y\left(s\right)=\frac{X\left(s\right)}{s+k}+\frac{y\left(0\right)}{s+k}$
Taking inverse Laplace transform , we have
$y\left(t\right)=e^{-kt}x\left(t\right)+y\left(0\right)e^{-kt}$
So, if we want $-2y\left(t\right)$ as a solution both $x\left(t\right)$ and $y\left(0\right)$ has to be multiplied by $-2$ ; hence change $x\left(t\right)$ by $-2x\left(t\right)$ and $y\left(0\right)$ by $-2y\left(0\right)$