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Problems #12, pg.164 and #23,pg.174 are graded.

**#12,pg.164:** Find the general solution of the given differential equation: $4y''+9y=0$.

**Solution:** This problem has characteristic equation
$$4r^2+9=0,$$
and so $r = \pm \sqrt{ -\dfrac{9}{4}} = \pm \dfrac{3}{2}i$. Therefore the general solution is
$$y(t)=c_1 \cos \left( \dfrac{3}{2}t \right) + c_2 \sin \left( \dfrac{3}{2} t \right).$$

**#23,pg.174:** Use the method of reduction of order to find a second solution of the given differential equation: $t^2y''-4ty'+6y=0, t>0$ given that $y_1(t)=t^2$ is a solution.

**Solution:** We assume that $y_2(t)=v(t)y_1(t)=v(t)t^2$. Then $y_2'(t)=2tv(t)+t^2v'(t)$ and
$$\begin{array}{ll}
y_2''(t)&=2v(t)+2tv'(t) + 2tv'(t)+t^2v''(t) \\
&= 2v(t) + 4tv'(t) + t^2v''(t).
\end{array}$$
Plug $y_2$ into the differential equation to get
$$t^2(2v(t)+4tv'(t)+t^2v''(t))-4t(2tv(t)+t^2v'(t))+6v(t)t^2=0.$$
Simplfying, we get
$$t^4 v''(t)=0,$$
but since $t>0$ we can divide by $t^4$ yielding
$$v''(t)=0.$$
Integrating this equation twice yields
$$v(t)=At+B.$$
This implies
$$y_2(t)=v(t)y_1(t)=(At+B)t^2 = At^3+Bt^2.$$
Therefore we see that the general solution is
$$y(t)=c_1 t^2 + c_2t^3.$$