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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 $$y(t)=c_1 t^2 + c_2t^3.$$