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# System prepended metadata

title: Homework 3
tags: [微分方程特論]

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# Homework 3 
ntu_b05202054 何信佑
###### tags: `微分方程特論`
紅筆訂正在 https://hackmd.io/@ulynx/ry5vBzmd8
## Section 5.1
### Problem 5
Transform the differential equation $$x^{(3)}=(x')^2+\cos x\tag{1}$$ into an equivalent system of first-order differential equations.
#### Solution 
Letting $x_1=x$, $x_2=x'$, $x_3=x''$, we have two equations $x_1'=x_2$ and $x_2'=x_3$.

Combining these two equations with Equation (1) in terms of $x_1$, $x_2$ and $x_3$, we can transform Equation (1) into $$\left\{\begin{array}{l}x_1'=x_2\\x_2'=x_3
\\x_3'=x_2^2+\cos x_1\end{array}\right..\blacksquare$$

### Problem 8
Transform the system of differential equations $$\left\{\begin{array}{l}x''+3x'+4x-2y=0\\y''+2y'-3x+y=\cos t\end{array}\right.\tag{1}$$ into an equivalent system of first-order differential equations.
#### Solution 
Letting $x_1=x$, $x_2=x'$, $x_3=y$, $x_4=y''$, we have two equations $x_1'=x_2$ and $x_3'=x_4$.

Combining these two equations with both of System (1) in terms of $x_1$, $x_2$, $x_3$ and $x_4$, we can transform System (1) into $$\left\{\begin{array}{l}x_1'=x_2\\x_3'=x_4\\x_2'=-3x_2-4x_1+2x_3\\x_4'=-2x_4+3x_1-x_3+\cos t\end{array}\right..\blacksquare$$

## Section 5.2
### Problem 21
Suppose that $L_1=a_1D^2+b_1D+c_1$ and $L_2=a_2D^2+b_2D+c_2$, where the coefficients are all constants, and that $x(t)$ is a ~~twice~~ <font color=red>fourth</font> differentiable function. Verify that $L_1L_2x=L_2L_1x$. 
#### Proof
$\require{color}\require{cancel}\begin{align}L_1L_2x&=(a_1D^2+b_1D+c_1)(a_2D^2+b_2D+c_2)x\\&=(a_1D^2+b_1D+c_1)(a_2x''+b_2x\color{red}{'}\color{black}+c_2x)\\&=a_1a_2x^{(4)}+b_1a_2x'''+c_1a_2x''\\&\quad+a_1b_2x'''+b_1b_2x''+c_1b_2x'\\&\quad+a_1c_2x''+b_1c_2x'+c_1c_2x\end{align}\tag{1}$

$\begin{align}L_2L_1x&=(a_2D^2+b_2D+c_2)(a_1D^2+b_1D+c_1)x\\&=(a_2D^2+b_2D+c_2)(a_1x''+b_1x\color{red}{'}\color{black}+c_1x)\\&=a_2a_1x^{(4)}+b_2a_1x'''+c_2a_1x''\\&\quad+a_2b_1x'''+b_2b_1x''+c_2b_1x'\\&\quad+a_2c_1x''+b_2c_1x'+c_2c_1x\end{align}\tag{2}$

Comparing each term of both (1) and (2), one can verify $L_1L_2x=L_2L_1x$. $\blacksquare$

### Problem 22
Suppose that $L_1x=tDx+x$ and that $L_2x=Dx+tx$. Show that $L_1L_2x\neq L_2L_1x$. Thus linear operators with *variable* coefficients generally do not commute.

#### Proof
$\begin{align}L_1L_2x&=(tD+1)(D+t)x\\&=(tD+1)(x'+tx)\\&=tx''+x'+t(x+\color{red}{\xcancel{2x'}+tx'}\color{black})+tx\\&=tx''+(\color{red}{\xcancel{2}}\color{black}{t}\color{red}{^2}\color{black}+1)x'+2tx\end{align}\tag{1}$

$\begin{align}L_2L_1x&=(D+t)(tD+1)x\\&=(D+t)(tx'+x)\\&=x'+tx''+t^2x'+x'+tx\\&=tx''+(t^2+2)x'+tx\end{align}\tag{2}$

Comparing each term of both (1) and (2), one can cleary see that $L_1L_2x\neq L_2L_1x$ in general. $\blacksquare$
<font color=red>Actually, the equality holds when $x(t)=ce^{t^2/2}$, $c$ being constant.</font>

## Section 5.3
### Problem 13
Write the following system in the form $\mathbf{x}'=\mathbf{P}(t)\mathbf{x}+\mathbf{f}(t)$:$$\left\{\begin{array}{l}x'=2x+4y+3e^t\\y'=5x-y-t^2\end{array}\right.\tag{1}$$
#### Solution
First we rewrite (1) in terms of $x_1=x$ and $x_2=y$:$$\left\{\begin{array}{l}x_1'=2x_1+4x_2+3e^t\\x_2'=5x_1-x_2-t^2\end{array}\right..\tag{2}$$And then we define $\mathbf{x}=\begin{pmatrix}x_1& x_2\end{pmatrix}^T$, so that $\mathbf{x}'=\begin{pmatrix}x_1'& x_2'\end{pmatrix}^T$. By doing so, System (2) can be regarded as $$\begin{pmatrix}x_1'\\x_2'\end{pmatrix}=\begin{pmatrix}2x_1+4x_2\\5x_1-x_2\end{pmatrix}+\begin{pmatrix}3e^t\\-t^2\end{pmatrix}.$$By further definition $\mathsf{P}(t)=\begin{pmatrix}2&4\\5&-1\end{pmatrix}$ and $\mathbf{f}(t)=\begin{pmatrix}3e^t\\-t^2\end{pmatrix}$, we recognize that $$\underbrace{\begin{pmatrix}x_1'\\x_2'\end{pmatrix}}_{\mathbf{x}'}=\underbrace{\begin{pmatrix}2&4\\5&-1\end{pmatrix}}_{\mathsf{P}(t)}\underbrace{\begin{pmatrix}x_1\\x_2\end{pmatrix}}_{\mathbf{x}}+\underbrace{\begin{pmatrix}3e^t\\-t^2\end{pmatrix}}_{\mathbf{f}(t)},$$ which is in accordance with the requested form. Actually, $\mathsf{P}(t)$ is independent of $t$.$\blacksquare$

### Problem 21
Given a system of equations $\mathbf{x}'=\begin{pmatrix}4&2\\-3&-1\end{pmatrix}\mathbf{x}$ and two vectors $\mathbf{x}_1=\begin{pmatrix}2e^t\\-3e^t\end{pmatrix}$ and $\mathbf{x}_2=\begin{pmatrix}e^{2t}\\-e^{2t}\end{pmatrix}$, 
1. verify the given vectors are solutions of the given system; 
2. use the Wronskian to show that they are linearly independent;
3. write the general solution of the system.
#### Solution
1. Substituting the vectors into the right hand side of the system, we find $$\begin{pmatrix}4&2\\-3&-1\end{pmatrix}\begin{pmatrix}2e^t\\-3e^t\end{pmatrix}=\begin{pmatrix}4\cdot 2e^t+2\cdot(-3)e^t\\-3\cdot 2e^t+(-1)(-3)e^t\end{pmatrix}=\begin{pmatrix}2e^t\\-3e^t\end{pmatrix}$$ and $$\begin{pmatrix}4&2\\-3&-1\end{pmatrix}\begin{pmatrix}e^{2t}\\-e^{2t}\end{pmatrix}=\begin{pmatrix}4e^{2t}+2\cdot(-1)e^{2t}\\-3e^{2t}+(-1)(-1)e^{2t}\end{pmatrix}=\begin{pmatrix}2e^{2t}\\-2e^{2t}\end{pmatrix},$$ which are exactly $\mathbf{x}_1'$ and $\mathbf{x}_2'$, respectively. Thus, the given vectors $\mathbf{x}_1$ and $\mathbf{x}_2$ are solutions of the system.
2. We can evaluate the Wronskian of $\mathbf{x}_1$ and $\mathbf{x}_2$: $$W(\mathbf{x}_1,\mathbf{x_2};t)=\begin{vmatrix}2e^t&-3e^t\\e^{2t}&-e^{2t}\end{vmatrix}=(2e^t)(-e^{2t})-(e^{2t})(-3e^t)=e^{3t}.$$ But since $e^{3t}\geq 0$ for any $t\in\mathbb{R}$, the vectors $\mathbf{x}_1$ and $\mathbf{x}_2$ are linearly independent.
<font color=red>I misunderstood the indices of entries in Wronskian (the Equation (33) of Section 5.3 in our textbook), changing rows to columns, although this does not alter the value of determinant. The textbook says: "Note that $W$ is the determinant of the matrix that has as its column vectors the solutions $\mathbf{x}_1, \mathbf{x}_2,\ldots, \mathbf{x}_n$."</font>
3. By the theory of linear system of equations, the general solution of $$\mathbf{x}'=\begin{pmatrix}4&2\\-3&-1\end{pmatrix}\mathbf{x}$$ is the linear combination of two linearly independent solutions $\mathbf{x}_1$ and $\mathbf{x}_2$, that is, $$\mathbf{x}=c_1\mathbf{x}_1+c_2\mathbf{x}_2=c_1\begin{pmatrix}2e^t\\-3e^t\end{pmatrix}+c_2\begin{pmatrix}e^{2t}\\-e^{2t}\end{pmatrix}=\begin{pmatrix}2c_1e^t+c_2e^{2t}\\-3c_1e^t-c_2e^{2t}\end{pmatrix},$$where $c_1$ and $c_2$ are arbitrary real constants. $\blacksquare$