Toronto Math Forum

$$\mbox{Find a particular solution of the given non homogeneous equation.}$$
$$t^2{y}''+7t{y}'+5y=t, \quad t>0; \quad y_{1}(t)=t^{-1} $$
$$\mbox{Let} \ y=v(t)y_{1}(t)$$
$$\mbox{Then} \ {y}'={v}'t^{-1}-vt^{-2}$$
$${y}''={v}''t^{-1}-{v}'t^{-2}-{v}'t^{-2}+2vt^{-3}$$
$$\mbox{Plug y, y' and y'' in the given non homogeneous equation}$$
$$\mbox{Then} \ t^{2}({v}''t^{-1}-{v}'t^{-2}-{v}'t^{-2}+2vt^{-3})+7t({v}'t^{-1}-vt^{-2})+5vt^{-1}=t{v}''+5{v}'$$
$${v}''+5{v}'t^{-1}=1$$
$$\mbox{Let} \ r={v}', \ {r}'={v}''$$
$${r}'+5{r}'t^{-1}=1$$
$$p(t)=5t^{-1}$$
$$\mu=e^{\int\frac{5}{t} dt}=t^{5}$$
$$t^{5}{r}'+5t^{4}r=t^{5}$$
$$r=\frac{1}{6}t+c_{1}t^{-5}={v}'$$
$$\mbox{Then} \ v=\int\frac{1}{6}t+c_{1}t^{-5}dt$$
$$v=\frac{1}{12}t^2-\frac{1}{4}c_{1}t^{-4}+c_{2}$$
$$\mbox{Thus,} \ y=\frac{1}{12}t^{2}t^{-1}-\frac{1}{4}c_{1}t^{-4}t^{-1}+c_{2}t^{-1}$$
$$y=\frac{1}{12}t-\frac{c_{1}}{4}t^{-5}+c_{2}t^{-1}$$
$$\mbox{Since} \ \frac{-c_{1}}{4} \mbox{is a constant. Therefore, the particular solution is } \ y=\frac{1}{12}t+c_{1}t^{-5}+c_{2}t^{-1}$$

$$\mbox{Find the solution of the given initial value problem }$$
$${2y}''+{y}'-4y=0$$
$$y\left(0 \right)=1 \mbox{ and } {y}'\left(0 \right)=0$$
$$2r^2+r-4=0$$
$$r=\frac{-1\pm\sqrt{1-4\times2\times(-4)}}{4}=\frac{-1\pm\sqrt{33}}{4}$$
$$\mbox{Hence, }r_{1}=\frac{-1+\sqrt{33}}{4} \mbox{ and }  r_{2}=\frac{-1-\sqrt{33}}{4}$$
$$\mbox{Then the general solution of the given differential equation is }$$
$$y=c_{1}e^{\frac{-1+\sqrt{33}}{4} t}+c_{2}e^{\frac{-1-\sqrt{33}}{4} t}$$
$$\because y\left(0 \right)=1$$
$$\therefore c_{1}+c_{2}=1$$
$${y}'=\frac{-1+\sqrt{33}}{4}c_{1}e^{\frac{-1+\sqrt{33}}{4} t}+\frac{-1-\sqrt{33}}{4}c_{2}e^{\frac{-1-\sqrt{33}}{4} t}$$
$$\because {y}'\left(0 \right)=0$$
$$\therefore \frac{-1+\sqrt{33}}{4}c_{1}+\frac{-1-\sqrt{33}}{4}c_{2}=0$$
$$\mbox{Since }c_{1}+c_{2}=1, \mbox{so }c_{1}=1-c_{2}$$
$$\therefore \frac{-1+\sqrt{33}}{4}(1-c_{2})+\frac{-1-\sqrt{33}}{4}c_{2}=0$$
$$ \frac{-1+\sqrt{33}}{4}-c_{2}\frac{2\sqrt{33}}{4}=0$$
$$\therefore c_{2}=\frac{33-\sqrt{33}}{66}$$
$$c_{1}=1-\frac{33-\sqrt{33}}{66}=\frac{33+\sqrt{33}}{66}$$
$$\mbox{Therefore, the solution of the given initial value problem is }$$
$$y=\frac{33+\sqrt{33}}{66}e^{\frac{-1+\sqrt{33}}{4} t}+\frac{33-\sqrt{33}}{66}e^{\frac{-1-\sqrt{33}}{4} t}$$