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**Quiz-1 / MOVED: MAT24f4 TUT5103 Quiz2**

« **on:**October 07, 2019, 05:32:24 AM »

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3

Answering question on discussions

Definitely the role of the Existence and Uniqueness Theorems are much more important than Limits in Manual Computations (and honestly, I have no idea what "Manual Computation" means). However this role is more theoretical both in ODEs and PDEs. Still, if we are talking about numerical solutions (taught in different classes, we skip Chapter 8, and briefly look at section 2.7) we need to be pretty sure that the object we are trying to find exist and we find all of them.

For centuries from I. Newton (who introduced ODEs) mathematicians did not care much about existence, because they were looking for solutions of real life problems and believed in existence and also because the rigorous apparatus of Real Analysis which allows to prove such theorems came into existence only in 19th century. You may want to look at very sketchy Lecture_Note_to_Section_2.8_Existence-Uniqueness_Theorem.

Uniqueness is a different matter: mathematicians observed that the solution to the Cauchy problem is not necessarily unique (remember, that the*general solution* to the 1st order ODE is $x=\varphi(t;C)$ or $\Phi(x, t; C)=0$ in the explicit and implicit form correspondingly and we need to specify one solution one needs to impose an extra condition; f.e. $x(t_0)=x_0$. They discovered that there could be a *singular solution* which is not a *regular solution* which means that it cannot be obtained from the general solution by freezing $C$ but which in each point coincides with some (depending on the point) regular solution. In more details see Lecture_Note_to_Chapter_2_Singular_Solutions.

Both of these lecture notes are optional

Definitely the role of the Existence and Uniqueness Theorems are much more important than Limits in Manual Computations (and honestly, I have no idea what "Manual Computation" means). However this role is more theoretical both in ODEs and PDEs. Still, if we are talking about numerical solutions (taught in different classes, we skip Chapter 8, and briefly look at section 2.7) we need to be pretty sure that the object we are trying to find exist and we find all of them.

For centuries from I. Newton (who introduced ODEs) mathematicians did not care much about existence, because they were looking for solutions of real life problems and believed in existence and also because the rigorous apparatus of Real Analysis which allows to prove such theorems came into existence only in 19th century. You may want to look at very sketchy Lecture_Note_to_Section_2.8_Existence-Uniqueness_Theorem.

Uniqueness is a different matter: mathematicians observed that the solution to the Cauchy problem is not necessarily unique (remember, that the

Both of these lecture notes are optional

4

In two examples were made errors.

**A.** Consider equation

\begin{equation}

y'= -\frac{y}{x}+y^2.

\label{1}

\end{equation}

It is*Bernoulli equation*. We solve it by the *method of variation of the constant*. Consider first corresponding linear homogeneous equation

\begin{equation}

y'= -\frac{y}{x}.

\label{2}

\end{equation}

It has a solution

\begin{equation}

y= Cx^{-1}

\label{3}

\end{equation}

with constant $C$ (do it by yourself!). Now consider (\ref{3}) with $C$ which is not a constant (variation!) and plug it into (\ref{1}).

\begin{gather*}

(Cx^{-1})' =-x^{-1}(Cx^{-1})+(Cx^{-1})^2\implies C'x^{-1}-Cx^{-2}=-Cx^{-2} + C^2x^{-2}\implies C' =C^2x^{-1}\\

\implies \frac{dC}{C^2}=\frac{dx}{x}\implies -\frac{1}{C}=\ln(x)+c

\end{gather*}

where $c=\mathsf{const}$. Then $C=-\dfrac{1}{\ln (x)+c}$ and plugging into (\ref{3}) we get

\begin{equation*}

\boxed{ y=-\frac{1}{x(\ln (x)+c)}.}

\end{equation*}

**B.** Consider equation

\begin{equation}

y'- \tan(x) y =\cos(x).

\label{4}

\end{equation}

We solve it by*the method of integrating factor*. Multiplying (\ref{4}) by unknown yet factor $\mu=\mu(x)$ we get

\begin{equation}

\mu y' - \mu \tan (x) y = \mu \cos(x).

\label{5}

\end{equation}

*We want* the left hand expression to be $\mu y'+\mu 'y$, which means

\begin{equation*}

\mu'=-\mu \tan(x) \implies \frac{d\mu}{\mu}= \tan(x)\,dx \implies \ln (\mu) =-\int \tan(x)\,dx = -\int\frac{\sin(x)}{\cos(x)}\,dx = \ln (\cos(x)).

\end{equation*}

You**must** know this integral. We do not need any constant since we need just one integrating factor.

So $\mu =\cos(x)$. (\ref{5}) is now

\begin{gather*}

\cos(x)y'-\sin (x)y =\cos^2(x)\implies \bigl(\cos(x)y\bigr)'= \cos^2(x)\implies

\cos(x)y = \int \cos^2(x)\,dx = \int \frac{1+\cos(2x)}{2}\,dx =\\

\frac{x}{2}+\frac{\sin (2x)}{4}+C=

\frac{x}{2}+\frac{\sin (x)\cos(x)}{2}+C.

\end{gather*}

You**must** know simple trigonometric formulae. Then

\begin{equation*}

\boxed{y= \frac{1}{2}(x+2C)\sec(x)+ \frac{1}{2}\sin (x).}

\end{equation*}

\begin{equation}

y'= -\frac{y}{x}+y^2.

\label{1}

\end{equation}

It is

\begin{equation}

y'= -\frac{y}{x}.

\label{2}

\end{equation}

It has a solution

\begin{equation}

y= Cx^{-1}

\label{3}

\end{equation}

with constant $C$ (do it by yourself!). Now consider (\ref{3}) with $C$ which is not a constant (variation!) and plug it into (\ref{1}).

\begin{gather*}

(Cx^{-1})' =-x^{-1}(Cx^{-1})+(Cx^{-1})^2\implies C'x^{-1}-Cx^{-2}=-Cx^{-2} + C^2x^{-2}\implies C' =C^2x^{-1}\\

\implies \frac{dC}{C^2}=\frac{dx}{x}\implies -\frac{1}{C}=\ln(x)+c

\end{gather*}

where $c=\mathsf{const}$. Then $C=-\dfrac{1}{\ln (x)+c}$ and plugging into (\ref{3}) we get

\begin{equation*}

\boxed{ y=-\frac{1}{x(\ln (x)+c)}.}

\end{equation*}

\begin{equation}

y'- \tan(x) y =\cos(x).

\label{4}

\end{equation}

We solve it by

\begin{equation}

\mu y' - \mu \tan (x) y = \mu \cos(x).

\label{5}

\end{equation}

\begin{equation*}

\mu'=-\mu \tan(x) \implies \frac{d\mu}{\mu}= \tan(x)\,dx \implies \ln (\mu) =-\int \tan(x)\,dx = -\int\frac{\sin(x)}{\cos(x)}\,dx = \ln (\cos(x)).

\end{equation*}

You

So $\mu =\cos(x)$. (\ref{5}) is now

\begin{gather*}

\cos(x)y'-\sin (x)y =\cos^2(x)\implies \bigl(\cos(x)y\bigr)'= \cos^2(x)\implies

\cos(x)y = \int \cos^2(x)\,dx = \int \frac{1+\cos(2x)}{2}\,dx =\\

\frac{x}{2}+\frac{\sin (2x)}{4}+C=

\frac{x}{2}+\frac{\sin (x)\cos(x)}{2}+C.

\end{gather*}

You

\begin{equation*}

\boxed{y= \frac{1}{2}(x+2C)\sec(x)+ \frac{1}{2}\sin (x).}

\end{equation*}

5

The main purpose of this forum is not the communication with instructors but peer-to-peer. I noticed that some of you answer the questions of your peers and I really appreciate, but I would like to see more -- much more -- of this!

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- This semester
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6

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7

Calculate for real $n>1$

$$

I:= \int_0^\infty\frac{dx}{1+x^n}.

$$

*Hint:* Consider

$$

\int_\gamma \frac{dz}{1+z^n}

$$

with with an arc of radius $R\to \infty$ and an angle $\alpha=\frac{2\pi}{n}$. Express the integral over the second straight segment through integral over the first one.

$$

I:= \int_0^\infty\frac{dx}{1+x^n}.

$$

$$

\int_\gamma \frac{dz}{1+z^n}

$$

with with an arc of radius $R\to \infty$ and an angle $\alpha=\frac{2\pi}{n}$. Express the integral over the second straight segment through integral over the first one.

8

Consider $P(z)= z^3 +2z -3-i$ and, using the argument theorem and RouchÃ©'s theorem calculate the number of its roots in each of the following domains:

**(a)** $\{z\colon |z-1|<1\}$;

**(b)** $\{z\colon |z-1|>1, |z|<2\}$,

**(c)** $\{z\colon |z|>2\}$.

9

10

Find all singular points, classify them, and find residues at these points of

$$

f(z)= \tan (z) + z\cot^2(z);

$$

infinity included.

$$

f(z)= \tan (z) + z\cot^2(z);

$$

infinity included.

11

$\{w=u+iv\colon \frac{u^2}{a^2}+\frac{v^2}{b^2}=1\}$ with $a^2-b^2=1$ and find $a=a(r)$ and $b=b(r)$.

$\{w=u+iv\colon \frac{u^2}{A^2}-\frac{v^2}{B^2}=1\}$ with $A^2+B^2=1$ and find $A=A(\theta)$ and $B=B(\theta)$.

12

Determine if the series is converging at $|z|=R$ (consider all points $R$ satisfying $|z|=R$).

13

For the system of ODEs

\begin{equation*}

\left\{\begin{aligned}

&x' = 2y(x^2+y^2+4)\, , \\

&y' = -2x (x^2+y^2-16)

\end{aligned}\right.

\end{equation*}

14

For the system of ODEs

\begin{equation*}

\left\{\begin{aligned}

&x' = x(3x +2y -30)\, , \\

&y' = y(2y-x-6)\,.

\end{aligned}\right.

\end{equation*}

15

Find the general solution $(x(t),y(t))$ of the system of ODEs

\begin{equation*}

\left\{\begin{aligned}

&x' = x-2y + \sec(t)\, &&-\frac{\pi}{2}<t<\frac{\pi}{2},\\

&y' = x -\ \,y \,.

\end{aligned}\right.

\end{equation*}