Solutions - 0265-03

(a)

First choose a function $x(t)$, then set $y(t)=3x(t)-4$ to ensure the equation is satisfied.

When choosing $x(t)$, we want $x$ to cover the whole domain of $y=3x-4$ which is $x\in (-\infty ,\infty )$. We also need $x(0)=2$ to satisfy the initial condition.

Start by trying $x(t)=t$:

$$y=3t-4$$

But then $x(0)=-4$. Since $c(0)=(\mathrm{2,2})$ we should have $x(0)=2$. We can arrange for this by setting $x(t)=t+a$ and solving for $a$:

$$x(0)=2\gg \gg 0+a=2\gg \gg a=2$$

Therefore we define $x(t)=t+2$. Then:

$$y(t)=3(t+2)-4\gg \gg 3t+2$$

So we use:

$$\begin{array}{cc}x(t)& =t+2\\ y(t)& =3t+2\end{array}$$

---

(b)

Same method but different condition:

$$x(3)=2\gg \gg 3+a=2\gg \gg a=-1$$

Therefore we define $x(t)=t-1$. Then:

$$y(t)=3(t-1)-4\gg \gg 3t-7$$

So we use:

$$\begin{array}{cc}x(t)& =t-1\\ y(t)& =3t-7\end{array}$$