Toronto Math Forum

$\textbf{Problem1:} \\$
$\textbf{(a)} \text{  Find the general solution of }$ $$x'=\begin{bmatrix}
13 & -9\\
6 & -8
\end{bmatrix}x$$
$\text{classify fixed point (0, 0) and sketch trajectories.} \\$
$\textbf{(b)} \text{  Find the general solution}$ $$x'=\begin{bmatrix}
13 & -9\\
6 & -8
\end{bmatrix}x + \begin{bmatrix}
0\\
\frac{150e^{25t}}{e^{30t}+1}
\end{bmatrix}x$$


$\textbf{Solution:} \\$
$\textbf{(a)}$

$\text{Let }$ $$A=\begin{bmatrix}
13 & -9\\
6 & -8
\end{bmatrix}$$

$\text{Then }$ $$det(A-\lambda I) = (13-\lambda)(-8-\lambda)-6\cdot(-9)=0$$

$\text{Solve for } \lambda: $ $$\lambda_1 = 10 , \lambda_2 = -5$$

$\text{When }\lambda = 10,$ $$A-\lambda I = \begin{bmatrix}
3 & -9\\
6 & -18
\end{bmatrix} \xrightarrow{\text{ref}}\begin{bmatrix}
1 & -3\\
0 & 0
\end{bmatrix}$$
$\text{Hence the eigenvector is }$ $$\vec{v_1}=\begin{bmatrix}
3\\
1
\end{bmatrix}$$

$\text{When }\lambda = -5,$ $$A-\lambda I = \begin{bmatrix}
18 & -9\\
6 & -3
\end{bmatrix} \xrightarrow{\text{ref}}\begin{bmatrix}
2 & -1\\
0 & 0
\end{bmatrix}$$
$\text{Hence the eigenvector is }$ $$\vec{v_2}=\begin{bmatrix}
1\\
2
\end{bmatrix}$$

$\text{Therefore we have }$ $$x=c_1e^{10t}\begin{bmatrix}
3\\
1
\end{bmatrix} + c_2e^{-5t}\begin{bmatrix}
1\\
2
\end{bmatrix}$$

$\text{(graph is attached below)}$


$\textbf{(b)}$

$\text{Let }$ $$\varphi (t)=\begin{bmatrix}
3e^{10t} & e^{-5t}\\
e^{10t} & 2e^{-5t}
\end{bmatrix}, u'(t) = \begin{bmatrix}
u_1'(t)\\
u_2'(t)
\end{bmatrix}$$

$\text{Thus we have }$ $$\left[ \begin{array}{cc|c}
3e^{10t} & e^{-5t} & 0 \\
e^{10t} & 2e^{-5t} & \frac{150e^{25t}}{e^{30t}+1}
\end{array}\right] \xrightarrow{R_2=3R_2-R_1} \left[ \begin{array}{cc|c}
3e^{10t} & e^{-5t} & 0 \\
0 & 5e^{-5t} & \frac{450e^{25t}}{e^{30t}+1}
\end{array}\right]$$

$\text{Observe the second row, we can get }$
$
\begin{gather}
\begin{aligned}

5e^{-5t}\cdot u_2'(t) &= \frac{450e^{25t}}{e^{30t}+1} \\ \\
u_2'(t) &= \frac{90e^{30t}}{e^{30t}+1}\\ \\
u_2(t) &=\int{\frac{90e^{30t}}{e^{30t}+1}} \,dt \\ \\
&= 90\int{\frac{e^{30t}}{e^{30t}+1}} \,dt \ \ \ \text{(integral by substitution)}\\ \\
&= 3ln(e^{30t}+1)+c_2

\end{aligned}
\end{gather}
$

$\text{Then compute the first row}$
$
\begin{gather}
\begin{aligned}

3e^{10t}u_1'(t)+e^{-5t}u_2'(t) &= 0 \\ \\
3e^{10t}u_1'(t)+e^{-5t}\frac{90e^{30t}}{e^{30t}+1} &= 0 \\ \\
u_1 &= -30\int{\frac{e^{15t}}{e^{30t}+1}} \,dt \ \ \ \text{(integral by substitution)}\\ \\
&=-30\int{\frac{\frac{1}{15}}{u^2+1}} \,du \\ \\
&= -2arctan(e^{15t})+c_1

\end{aligned}
\end{gather}
$

$\text{Therefore }$
$
\begin{gather}
\begin{aligned}

x&=\varphi{(t)}\cdot u(t) \\ \\
&= \begin{bmatrix}
3e^{10t} & e^{-5t}\\
e^{10t} & 2e^{-5t}
\end{bmatrix} \cdot \begin{bmatrix}
-2arctan(e^{15t})+c_1\\
3ln(e^{30t}+1)+c_2
\end{bmatrix} \\ \\
&= c_1\begin{bmatrix}
3e^{10t}\\
e^{10t}
\end{bmatrix} +c_2\begin{bmatrix}
e^{-5t}\\
2e^{-5t}
\end{bmatrix} + \begin{bmatrix}
-6e^{10t}arctan(e^{15t})+3e^{-5t}ln(e^{30t}+1)\\
-2e^{10t}arctan(e^{15t})+6e^{-5t}ln(e^{30t}+1)
\end{bmatrix}

\end{aligned}
\end{gather}
$

$\textbf{Problem2:} \\ \\
\text{Find the power series expansion at z=0 of}$
$$f(z) = \frac{1}{z} \int_{0}^{z}{\frac{1}{z}\int_{0}^{z}{cos(w^2)\,dw}\,dz}$$
$\text{What is the radius of convergence?}$

$\textbf{Solution:} \\ \\
\text{Firstly, we have power series,}$
$$cos(w) = \sum_{n=0}^{\infty}{(-1)^{n}\frac{w^{2n}}{(2n)!}} \\ $$
$\text{with a radius of convergence of } \infty \\$

$\text{Then substitute } w \text{ by } w^2$
$$cos(w^2) = \sum_{n=0}^{\infty}{(-1)^{n}\frac{w^{4n}}{(2n)!}}$$

$\text{Now substitute the series in the integral and calculate the integral}$

$
\begin{gather}
\begin{aligned}

f(z) &= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}} \int_{0}^{z}{cos(w^2)\,dw}\,dz \\\\
&= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}}(\int_{0}^{z} {\sum_{n=0}^{\infty}{\frac{(-1)^{n}w^{4n}}{(2n)!}\,}}\,dw) \,dz \\\\
&= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}}\sum_{n=0}^{\infty}{
\int_{0}^{z}{(-1)^{n} \cdot \frac{w^{4n}}{(2n)!}}\,dw}\,dz \\\\
&=\frac{1}{z} \int_{0}^{z}{\frac{1}{z}} \sum_{n=0}^{\infty}{(-1)^{n}\cdot \frac{z^{4n+1}}{(2n)! \cdot (4n+1)}}\,dz \\\\
&= \frac{1}{z} \int_{0}^{z}{\sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)}}} \,dz\\\\
&= \frac{1}{z} \sum_{n=0}^{\infty}{\int_{0}^{z}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)}}}\,dz \\\\
&= \frac{1}{z} \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n+1}}{(2n)! \cdot (4n+1)^{2}}} \\\\
&= \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)^{2}}}

\end{aligned}
\end{gather}
$

$\text{Hence the power series expansion is }$
$$f(z) = \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)^{2}}}$$

$\text{Since the radius of convergence of } cos(z) \text{ is } \infty \text{. Thus } f(z) \text{ has the same radius of convergence of  } R=\infty.$

$\textbf{Problem:} \\\\
\text{Compute the following integral}$ $$\int_{-\infty}^{\infty}{\frac{cos(x)}{(x+a)^{2}+b^2}} \,dx$$
$\text{with } b>0.$


$\textbf{Solution:} \\\\$
$\text{Let }$ $$f(z) = \frac{e^{iz}}{(z+a)^{2}+b^{2}}$$

$\text{Then we have two poles: } z_1 = -a-ib ,\ z_2 = -a+ib$

$\text{Since we are looking for upper-half plane, then we have pole at } z=-a+ib \text{ with order 1.}$

$\text{(Denote the path as the pic attached below)}$

$\text{By Residual Theorem we have: }$
$
\begin{gather}
\begin{aligned}

\int_{\gamma}{f(z)} \,dz &= 2 \pi i \cdot Res{f; -a+ib} \\\\
&=2 \pi i \cdot \frac{e^{iz}}{z+a+ib}|_{z=-a+ib} \\\\
&= \frac{\pi e^{-b} e^{-ia}}{b}

\end{aligned}
\end{gather}
$

$\text{For path } c_2:$

$
\begin{gather}
\begin{aligned}

|\int_{c_2}{f(z)} \, dz| &= |\int_{0}^{\pi}{f(Re^{i\theta})Re^{i\theta}} \,d\theta| \\\\
&= |\int_{0}^{pi} {\frac{e^{iRe^{i\theta}}}{(Re^{i\theta}+a)^{2}+b^{2}}} \,d\theta| \\\\
&\leq \pi R \cdot |max(f(z))| \\\\
&= \pi R \cdot max|\frac{e^{iRe^{i\theta}}}{(Re^{i\theta}+a)^{2}+b^{2}}| \\\\
&= \pi R \cdot max\frac{e^{|iR(cos(\theta)+isin(\theta))}|}{|(Re^{i\theta}+a)^{2}+b^{2}|}\\\\
&\leq \pi R \cdot \frac{e^{-Rsin(\theta)}}{|(R+a)^{2} - b^{2}|} \ \ \ \ \ \text{(since triangle inequality)} \\\\
&\leq \pi R \cdot \frac{e^{0}}{(R+a)^{2}+b^{2}} \\\\
&= 0  \ \ \text{as R} \rightarrow \infty \\\\
\end{aligned}
\end{gather}
$ $$\text{(By comparing the power of R on both numerator and denominator.)}$$

$\text{Now solve for path } c_1:$

$
\begin{gather}
\begin{aligned}
\int_{c_1}{f(z)}\,dz &= \int_{-\infty}^{\infty}{\frac{e^{ix}}{(x+a)^{2}+b^{2}}}\,dx \\\\
&= \int_{-\infty}^{\infty}{\frac{cos(x)}{(x+a)^{2}+b^{2}}}\,dx + i\int_{-\infty}^{\infty}{\frac{sin(x)}{(x+a)^{2}+b^{2}}}\,dx \\\\
\end{aligned}
\end{gather}
$
$\text{Since the imaginary part of the integral is an odd function, thus its integral is 0.}$

$\text{Thus we have:}$
$$\int_{-\infty}^{\infty}{\frac{cos(x)}{(x+a)^{2}+b^{2}}}\,dx = \frac{\pi e^{-b} e^{-ia}}{b}$$

$
\textbf{Problem2:} \\\\

\text{Find the general solution of }$
$$y'''+y' = \frac{2}{sin(2t)}$$
$\text{Hint: Use method of variations.} \int{sec(t)}\,dt = ln(tan(t)+sec(t)) \text{ and } \int{csc(t)}\,dt = -ln(cot(t)+csc(t))$

$\textbf{Solution: }$

$\text{Consider the characteristic polynomial:}$
$$\lambda^3 + \lambda = 0 \\\\
\lambda(\lambda^2+1) = 0 \\\\
\lambda_1 =0, \lambda_2=i, \lambda_3=-i$$
$\text{Hence we have }$
$$y_1 = c_1e^0 = c_1 \\\\
y_2 = c_2cos(t) \\\\
y_3 = c_3sin(t)
$$

$\text{Take} c_1 = c_2 = c_3 = 1 \text{, we have}$
$$ w_1 =\begin{bmatrix}
cos(t) & sin(t) \\
-sin(t) & cos(t) \\
\end{bmatrix} = 1$$

$$ w_2 = -\begin{bmatrix}
1 & sin(t) \\
0 & cos(t) \\
\end{bmatrix} = -cos(t)$$

$$ w_3 = \begin{bmatrix}
1 & cos(t) \\
0 & -sin(t) \\
\end{bmatrix} = -sin(t)$$

$\text{Then we have Wronskian of }y_1 ,\ y_2 ,\ y_3$
$
\begin{gather}
\begin{aligned}
W&= y_1''w_1 + y_2''w_2+y_3''w_3 \\\\
&= 0+ (cos(t))''(-cos(t)) + (sin(t))''(-sin(t)) \\\\
&= cos^2(t)+sin^2(t) \\\\
& =1
\end{aligned}
\end{gather}
$

$\text{To compute }u_1 ,\ u_2,\ u_3 \text{, we need to use the hind in integral computation} $

$
\begin{gather}
\begin{aligned}

u_1 &= \int{\frac{1\cdot \frac{2}{sin(2t)}}{1}} \, dt \\\\
&= 2 \int{csc(2t)} \,dt \ \ \ \text{ (by the hint)} \\\\
&= -2 \cdot ln(cot(2t)+csc(2t)) + c_1

\end{aligned}
\end{gather}
$


$
\begin{gather}
\begin{aligned}

u_2 &=\int{\frac{(-cos(t))\cdot \frac{2}{sin(2t)}}{1}} \, dt \\\\
& = -2\int{\frac{cos(t)}{sin(2t)}} \,dt \\\\
&= -2\int{\frac{cos(t)}{2sin(t)cos(t)}} \,dt \\\\
&= - \int{csc(t)} \,dt \ \ \ \text{ (by the hint)} \\\\
&= ln(cot(t) + csc(t)) + c_2

\end{aligned}
\end{gather}
$


$
\begin{gather}
\begin{aligned}

u_3 &=\int{\frac{(-sin(t))\cdot \frac{2}{sin(2t)}}{1}} \, dt \\\\
& = -2\int{\frac{sin(t)}{2sin(t)cos(t)}} \,dt \\\\
& = - \int{sec(t)} \,dt \ \ \ \text{ (by the hint)} \\\\
&= -ln(tan(t) + sec(t)) +c_3

\end{aligned}
\end{gather}
$

$\text{Therefore, the general solution is}$

$
\begin{gather}
\begin{aligned}

y &= u_1y_1 + u_2y_2 + u_3y_3 \\\\
&= -2 \cdot ln(cot(2t)+csc(2t)) + c_1 + cos(t)\cdot(ln(cot(t) + csc(t)) + c_2) + sin(t) \cdot (-ln(tan(t) + sec(t)) +c_3) \ \ \ \ \text{(by rearranging, we have)}\\\\
& =c_1 + c_2cos(t) + c_3sin(t) - 2 \cdot ln(cot(2t)+csc(2t)) + cos(t) \cdot ln(cot(t) + csc(t)) - sin(t) \cdot ln(tan(t) + sec(t)

\end{aligned}
\end{gather}
$

$
\textbf{For part (a):} \\\\
\text{By ratio test, we have : }\\\\
$

$
\begin{gather}
\begin{aligned}

\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| &= \lim_{n \to \infty}|\frac{(1+ i\sqrt{3})^{n+1}}{4^{n+1}(n+1)log^{3}(n+1)} \cdot \frac{n 4^{n} log^{3}(n)}{(1+i\sqrt{3})^{n}}| \\\\
&= \lim_{n \to \infty} |\frac{1+i\sqrt{3}}{4} \cdot \frac{nlog^{3}(n)}{(n+1)log^3(n+1)}| \\\\
&= \frac{1}{2} \text{(by further expanding this absolute value)}\\\\
&= \frac{1}{R}

\end{aligned}
\end{gather}
$

$\text{Therefore the radius of convergence is 2.}$


$
\textbf{For part (b):} \\\\
\text{We can follow the similar method. By ratio test, we have : }\\\\
$

$
\begin{gather}
\begin{aligned}

\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| &= \lim_{n \to \infty} |\frac{(i+\sqrt{3})^{n+1} \cdot \frac{e^{2(n)+1}-1}{e^{2(n+1)}+1}}{4^{n+1}} \cdot \frac{4^n}{(i+\sqrt{3})^{n}\cdot \frac{e^{2n}-1}{e^{2n}+1}}| \\\\
&= \lim_{n \to \infty} |\frac{i+\sqrt{3}}{4} \cdot \frac{e^{2n+2}-1}{e^{2n+2}+1} \cdot \frac{e^{2n}+1}{e^{2n}-1}| \\\\
&= \frac{1}{2} \text{(by further expanding this absolute value, and most terms can be canceled out.)}\\\\
&= \frac{1}{R}

\end{aligned}
\end{gather}
$

$\text{Therefore the radius of convergence is 2.} \\\\$

$\text{Above is how I got my answers. Correct me if I made any mistakes. Feel free to comment below }$

$
\textbf{Problem:} \\\\
\text{(a) Find the general solution for equation: } \ y'' -4y' + 5y = 30 + 40cos(3t) \\\\
\text{(b) Find solution that satisfies:} \ y(0) = 0, y'(0) = 0
$

$
\textbf{Solution for part (a): } \\\\
\text{Solve for homogenous solution:} \\\\

\begin{gather}
\begin{aligned}

r^{2} - 4r + 5 = 0 \Longrightarrow r = 2 \pm i \\\\

\end{aligned}
\end{gather}
$

$
\text{Therefore, } \\\\
\begin{gather}
\begin{aligned}

y_h = c_1 e^{2t}cos(t) + c_2 e^{2t}sin(t)

\end{aligned}
\end{gather}
$

$
\text{Let } y_p = At+B+Csin(3t)+Dcos(3t) \ \text{ , then we have} \\\\

\begin{gather}
\begin{aligned}

y'_p &= A+3Ccos(3t) -3Dsin(3t) \\\\
y''_p &= -9Csin(3t)-9Dcos(3t)

\end{aligned}
\end{gather}
$

$
\text{Plug them in the original equation, we have: } \\\\

\begin{gather}
\begin{aligned}

y'' -4y' + 5y &= -9Csin(3t)-9Dcos(3t) -4(A+3Ccos(3t) -3Dsin(3t)) + 5(At+B+Csin(3t)+Dcos(3t)) \\\\
&= 30 + 40cos(3t)

\end{aligned}
\end{gather}
$

$
\text{Solve for parameter A, B, C, and D, we get } A=0 ,\ B=6,\ C=-3,\ D=-1\\\\
\text{Thus we have: } \\\\ \\\\

\begin{gather}
\begin{aligned}

Y &= y_h + y_p \\\\
&= c_1 e^{2t}cos(t) + c_2 e^{2t}sin(t) -3sin(3t) -cos(3t) + 6

\end{aligned}
\end{gather}
$


$
\textbf{Solution for part (b): } \\\\
$

$
\text{Plug in } y(0)=0 \text{, we get } \ c_1 = -5 \text{, thus we have } \\\\
\begin{gather}
\begin{aligned}
Y = -5e^{2t}cos(t) + c_2 e^{2t}sin(t) -3sin(3t) -cos(3t) + 6
\end{aligned}
\end{gather}
$

$
\text{Now consider } y'(0)=0 \text{, then we have:}\\\\
\begin{gather}
\begin{aligned}
0 = -5(2e^{2t}cos(t) - e^{2t}sin(t)) + c_2(2e^{2t}sin(t)+e^{2t}cos(t)) - 9cos(3t) + 3sin(3t) \\\\
\Longrightarrow c_2 = 19
\end{aligned}
\end{gather}
$

$
\text{Thus the solution is:}\\\\

\begin{gather}
\begin{aligned}
Y = -5e^{2t}cos(t) + 19 e^{2t}sin(t) -3sin(3t) -cos(3t) + 6
\end{aligned}
\end{gather}
$

$
\textbf{Problem 1:} \\\\
\text{(a) Show that  } \ 3x^{4} - 18x^{2}y^{2} + 3y^{4} - 3x  \text{ is a harmonic function;} \\\\
\text{(b) Find the harmonic conjugate function } v(x,y) \\\\
\text{(c) Consider } u(x,y) + iv(x,y) \ \text{ and write it as a function } f(z) \text{ of } z=x+iy
$

$
\textbf{Solution for part (a): } \\\\

\text{Since we have:} \\\\
$
$
\begin{gather}
\begin{aligned}

U(x,y) = 3x^{4} - 18x^{2}y^{2} + 3y^{4} - 3x \\\\

\end{aligned}
\end{gather}
$

$
\text{then we have the following:} \\\\
$
$
\begin{gather}
\begin{aligned}

u_x &= 12x^{3} - 36xy^{2} - 3 \\\\
u_{xx} &= 36x^{2} - 36y^{2}\\\\
u_y &= -36x^{2}y + 12y^{3} \\\\
u_{yy} &= -36x^{2} + 36y{2}

\end{aligned}
\end{gather}
$

$\text{Thus we have: } \\\\$

$
\begin{gather}
\begin{aligned}

\Delta u &= u_{xx} + u_{yy} \\\\
&= 36x^{2} - 36y^{2} - 36x^{2} + 36y^{2} \\\\
&= 0

\end{aligned}
\end{gather}
$
$\text{Therefore the function is harmonic. } \\\\$



$
\\\\
\\\\
\textbf{Solution for part (b): }
\\\\

\text{We have:} \\\\
\begin{gather}
\begin{aligned}

v_x &= -u_y = 36x^{2}y - 12y^{3} \\\\
v_y &= u_x = 12x^{3} -36xy^{2} -3

\end{aligned}
\end{gather}
$

$
\text{Then} \\\\

\begin{gather}
\begin{aligned}

v_{x,y} &= \int{v_x} \,dx + \phi(y) \\\\
&= 12x^{3}y-12xy^{3}+\phi(y)

\end{aligned}
\end{gather}
$

$
\text{Now we calculate  } \phi(y) \text{ by following:} \\\\

\begin{gather}
\begin{aligned}

v_y &= (v_{x,y})' \\\\
&= 12x^{3} -36xy^{2}-3

\end{aligned}
\end{gather}
$

$
\text{thus, } \\\\

\begin{gather}
\begin{aligned}

\phi'(y) &= -3 \\\\
\phi(y) &= \int{-3}\,dy \\\\
&= -3y + c

\end{aligned}
\end{gather}
$

$
\text{Therefore} \\\\

\begin{gather}
\begin{aligned}

v_{x,y} = 12x^{3}y-12xy^{3}-3y+c
\end{aligned}
\end{gather}
$


$
\\\\
\\\\
\textbf{Solution for part (c): }
\\\\
\text{Let } \ z=x+iy \\\\

\begin{gather}
\begin{aligned}

u_{x,y} + iv_{x,y} &= 3x^{4} -12x^{2}y^{2} + 3y^{4} -3x +i(12x^{3}y-12xy^{3}-3y)+iC \\\\
&= 3(3x^{4} -12x^{2}y^{2} + 3y^{4} + i12x^{3}y-i12xy^{3}) -3(x+iy) +iC\\\\
&= 3(x+iy)^{4}-3(x+iy)+iC\\\\
&= 3z^{4} -3z + iC

\end{aligned}
\end{gather}
$