# Homework 2
ntu_b05202054 何信佑
紅筆訂正在 https://hackmd.io/@ulynx/BkLpG2gD8
###### tags: `微分方程特論`
$\require{color}\require{cancel}$
Given a system of differential equations $$\left\{\begin{matrix}\begin{array}{l}2\dfrac{dx}{dt} - 2\dfrac{dy}{dt} -3x =t\\2\dfrac{dx}{dt} +2\dfrac{dy}{dt} +3x +8y =2\end{array}\end{matrix},\right.\tag{1}$$with the operator $D\equiv\dfrac{d}{dt}$, we can write$$\left\{\begin{matrix}\begin{array}{l}2Dx - 2Dy -3x =t\\2Dx +2Dy +3x +8y =2\end{array}\end{matrix}\right.$$ or $$\left\{\begin{matrix}\begin{array}{l}(2D-3)x - 2Dy =t\\(2D+3)x +(2D+8)y =2\end{array}\end{matrix}\right..\tag{2}$$

Then, the operational determinant of this system is$$\begin{align}\color{red}{\Delta:=}\color{black} \begin{vmatrix}2D-3&-2D\\2D+3&2D+8\end{vmatrix}&=(2D-3)(2D+8)-(-2D)(2D+3)\\&=8D^2+16D-24.\end{align}$$

<font color=red>Also, we have</font> $$\color{red}{\begin{align}\Delta_x&:=\begin{vmatrix}t&-2D\\2&2D+8\end{vmatrix} =(2D+8)t-(-2D)2,\\\Delta_y&:=\begin{vmatrix}2D-3&t\\2D+3&2\end{vmatrix} =(2D+3)t+(2D-3)2.\end{align}}$$

Hence, by the procedure similar to the Cramer's rule, the system of equations can be rewritten as $$\left\{\begin{matrix}\begin{array}{l}(8D^2+16D-24)x=(2D+8)t-(-2D)2\\ (8D^2+16D-24)y=(2D+3)t+(2D-3)2\end{array}\end{matrix}\right.$$ or
$$\left\{\begin{matrix}\begin{array}{l}8x''+16x'-24x=8t+2\\ 8y''+16y'-24y=-3t-8\end{array}\end{matrix}\right.,\tag{3}$$ which is a pair of nonhomogeneous equations.

The characteristic equation of each of (3) is $$8r^2+16r-24=8(r-1)(r+3)=0,$$ so the general solutions of their associated homogeneous equations are $$\left\{\begin{matrix}\begin{array}{l}x(t)=a_1e^t+a_2e^{-3t}\\ y(t)=b_1e^t+b_2e^{-3t}\end{array}\end{matrix}\right.\tag{4}$$

Now we need to figure out the ~~complementary~~ <font color=red>particular</font> solutions. <font color=red>Since both of the nonhomogenrous terms of (4) are polynomials of degree 1, we can assume $x_p(t)=A_1t+A_2$ and $y_p(t)=B_1t+B_2$. (None of their terms appear in (4).)</font>

Substituting $x_p(t)=A_1t+A_2$, $x'_p(t)=A_1$, and $x''_p(t)=0$ into the first equation of (3), we have $$\begin{align}0+16A_1-24(A_1t+A_2)&=8t+2,\\(-24A_1)t+(16A_1-24A_2)&=8t+2,\end{align}$$ which yields $A_1=-\frac{1}{3}$ and $A_2=-\frac{11}{36}$.

Likewise, substituting $y_p(t)=B_1t+B_2$, $y'_p(t)=B_1$, and $y''_p(t)=0$ into the second equation of (3), we have $$\begin{align}0+16B_1-24(B_1t+B_2)&=-3t-8,\\(-24B_1)t+(16B_1-24B_2)&=-3t-8,\end{align}$$ which yields $B_1=\frac{1}{8}$ and $B_2=\frac{5}{12}$.

Therefore, the solutions of nonhomogeneous equations (3) are $$\left\{\begin{matrix}\begin{array}{l}x(t)=a_1e^t+a_2e^{-3t}-\frac{1}{3}t-\frac{11}{36}\\ y(t)=b_1e^t+b_2e^{-3t}+\frac{1}{8}t+\frac{5}{12}\end{array}\end{matrix}\right..\tag{5}$$

There are four *arbitrary* constants $a_1$, $a_2$, $b_1$, $b_2$ here. However, the order of the operational determinant is 2, which implies there only exists two arbitrary constants, with two relations in disguise.

To figure out the relations between them, one may substitute (5) into each of the original system of equations (1): $$2\left(a_1e^t-3a_2e^{-3t}-\dfrac{1}{3}\right)-2\left(b_1e^t-3b_2e^{-3t}+\dfrac{1}{8}\right)-3\left(a_1e^t+a_2e^{-3t}-\frac{1}{3}t-\frac{11}{36}\right)=t\\\left(2a_1-2b_1-3a_1\right)e^t+\left(-6a_2+6b_2-3a_2\right)e^{-3t}+\cancelto{0}{\left(\dfrac{3}{3}-1\right)}t+\cancelto{0}{\left(-\dfrac{2}{3}-\dfrac{2}{8}+\dfrac{33}{36}\right)}=0$$ Since $e^{t}$ and $e^{-3t}$ are linearly independent, the above identity requires that $a_1+2b_1=0$ and $-3a_2+2b_2=0$, or $b_1=-\frac{1}{2}a_1$ and $b_2=\frac{3}{2}a_2$.

It turns out that the solution of (1) is $$\left\{\begin{matrix}\begin{array}{l}x(t)=a_1e^t+a_2e^{-3t}-\frac{1}{3}t-\frac{11}{36}\\ y(t)=-\frac{1}{2}a_1e^t+\frac{3}{2}a_2e^{-3t}+\frac{1}{8}t+\frac{5}{12}\end{array}\end{matrix}\right..\tag{6}$$