Question

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

• A. change the initial condition to -y(0) and the forcing function to 2x(t)
• B. change the initial condition to 2y(0) and the forcing function to (t)
• 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(0) and the forcing function to -2x(t)

Answer 1

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

Consider first order differential equation is Linear form with constant coefficient

$\frac{dy}{dt}+ay=x\left(t\right)$

then, integrating factor

I.F = $e^{\int adt}$

& solution is

y (I.F) = $\int x\left(t\right).I.Fdt+c$

$\Rightarrow y=\frac{1}{I.F}\int x\left(t\right).IFdt+\frac{C}{IF}$

$\because$ C can be evaluated using initial condition so we can take y(0) = $\frac{C}{I.F}$

So, $y\left(t\right)=\frac{1}{I.F}\int x\left(t\right)I.Fdt+y\left(0\right)$

Multiply both sides with (-2)

or $-2y\left(t\right)=\frac{-2}{I.F}\int x\left(t\right)I.F.dt+(-2)y\left(0\right)$

$-2y\left(t\right)=\frac{1}{I.F}\int [-2x(t\left)\right]I.F.dt+[-2y(0\left)\right]$

Hence, for the solution to be -2y(t), we have to change x(t) by -2x(t) and the initial condition y(0) by -2y(0).