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Problems #1 and #7 from pg.144 are graded.

**#1,pg.144:** Find the general solution of $y''+2y'-3y=0$.

**Solution:** If we let $y=e^{rt}$ and substitue in, we get the characteristic equation
$$(*) \hspace{35pt} r^2+2r-3=0.$$
Since the left hand side factors to $(r+3)(r-1)$ we see the solution to $(*)$ is $r=1,-3$. Therefore the general solution is
$$y(t)=c_1 e^{t}+ c_2e^{-3t}.$$

**#7,pg.144:** Find the general solution of $y''-9y'+9y=0$.

**Solution:** The characteristic equation is
$$r^2-9r+9=0,$$
whose solution is given by the quadratic formula:
$$r= \dfrac{9 \pm \sqrt{81-4(9)}}{2} = \dfrac{9}{2} \pm \dfrac{\sqrt{45}}{2}.$$
Thus the general solution is
$$y(t) = c_1 e^{(\frac{9}{2} + \frac{\sqrt{45}}{2})t}+c_2 e^{(\frac{9}{2}-\frac{\sqrt{45}}{2})t}$$
Note: the solution given by Wolfram alpha looks different, but is algebraically equal. Neither one of these is "more correct" than the other.