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**Chapter 3 / Re: Confused about the notation in maximum modulus principle**

« **on:**December 16, 2020, 04:16:44 PM »

woww thank you, that makes sense now.

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woww thank you, that makes sense now.

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Hello, this is the problem 5 from the sample exam 2020F, I am kind of confused by the notation here. What is meant by find instead $max_D Re(f(z))$ and $\in_D Re(f(z))$? Are we suppose to find the maximum of the real part of the function f on the bounded domain D?

3

Questions: Find all points of continuity of the given function: $f(z)=(Imz-Rez)^{-1}$

Solutions:

Let $z=x+iy$ where x,y are real numbers.

Then $f(z)=f(x,y)=(Im(x+iy)-Re(x+iy))^{-1}=(y-x)^{-1}=\frac{1}{y-x}$

The function is not valid when the denominator equals 0, that is, $y=x$.

Therefore, the function is discontinuous only at all points of the line $y=x$.

Solutions:

Let $z=x+iy$ where x,y are real numbers.

Then $f(z)=f(x,y)=(Im(x+iy)-Re(x+iy))^{-1}=(y-x)^{-1}=\frac{1}{y-x}$

The function is not valid when the denominator equals 0, that is, $y=x$.

Therefore, the function is discontinuous only at all points of the line $y=x$.

4

Problem:

Describe the locus of points z satisfying the given equation.

$|z+1|^2+2|z|^2=|z−1|^2.$

Solution:

Let $z=x+yi$ where x,y are real numbers

$|z+1|^2 = |(x+yi)+1|^2=|(x+1)+yi|^2=(x+1)^2+y^2=x^2+2x+1+y^2$

$2|z|^2=2|x+yi|^2=2(x^2+y^2)=2x^2+2y^2$

$|z-1|^2 = |(x+yi)-1|^2=|(x-1)+yi|^2=(x-1)^2+y^2=x^2-2x+1+y^2$

Hence,

$|z+1|^2+2|z|^2=|z−1|^2$

$x^2+2x+1+y^2+2x^2+2y^2=x^2-2x+1+y^2$

$2x^2+4x+2y^2=0$

$2(x^2+2x+y^2)=0$

$x^2+2x+y^2=0$

$x^2+2x+1+y^2=0$

$(x+1)^2+y^2=1$

The locus of point z is a circle centered at (-1,0) with radius 1.

Describe the locus of points z satisfying the given equation.

$|z+1|^2+2|z|^2=|z−1|^2.$

Solution:

Let $z=x+yi$ where x,y are real numbers

$|z+1|^2 = |(x+yi)+1|^2=|(x+1)+yi|^2=(x+1)^2+y^2=x^2+2x+1+y^2$

$2|z|^2=2|x+yi|^2=2(x^2+y^2)=2x^2+2y^2$

$|z-1|^2 = |(x+yi)-1|^2=|(x-1)+yi|^2=(x-1)^2+y^2=x^2-2x+1+y^2$

Hence,

$|z+1|^2+2|z|^2=|z−1|^2$

$x^2+2x+1+y^2+2x^2+2y^2=x^2-2x+1+y^2$

$2x^2+4x+2y^2=0$

$2(x^2+2x+y^2)=0$

$x^2+2x+y^2=0$

$x^2+2x+1+y^2=0$

$(x+1)^2+y^2=1$

The locus of point z is a circle centered at (-1,0) with radius 1.

5

a) Find the general solution for equation $y'' - 2y' -3y = 16coshx$

b) Find solution, satisfying $y(0) = 0, y'(0) = 0$

a)

Consider $y'' - 2y' -3y = 0$

Assume $y = e^{rx}$, then the characteristic polynomial of the equation is given by $r^2 -2r -3 = 0$

This simplifies to $(r - 3)(r +1) = 0$ which gives us two roots $r_1 = 3$ and $r_2 = -1$

Then the complementary solution is $y_c = c_1e^{3x} + c_2e^{-x}$

Since $y'' - 2y' -3y = 16cosh = 8e^x + 8e^{-x}$

We guess the particular solution is of the form

$y = Ae^x + Bxe^{-x}$

$y' = Ae^x + B(e^{-x} - xe^{-x}) = Ae^x + Be^{-x} - Bxe^{-x}$

$y'' = Ae^x - Be^{-x} - B(e^{-x} - xe^{-x}) = Ae^x - 2Be^{-x} + Bxe^{-x}$

Substituting back to the original equation:

$y'' - 2y' -3y = Ae^x - 2Be^{-x} + Bxe^{-x} - 2Ae^x - 2Be^{-x} + 2Bxe^{-x} - 3Ae^x - 3Bxe^{-x} = -4Ae^x - 4Be^{-x} = 8e^x + 8e^{-x}$

Therefore, $A = -2$ and $B = -2$

The particular solution is $y_p = - 2e^x - 2xe^{-x}$

The general solution is :

$$y = c_1e^{3x} + c_2e^{-x} - 2e^x - 2xe^{-x}$$

b)

$$y = c_1e^{3x} + c_2e^{-x} - 2e^x - 2xe^{-x}$$

$$y' = 3c_1e^{3x} - c_2e^{-x} - 2e^x + 2xe^{-x} - 2e^{-x}$$

Plug in the initial conditions $y(0) = 0, y'(0) = 0$, we have,

$0 = c_1 + c_2 - 2$

$0 = 3c_1 - c_2 - 2 - 2$

so, $c_1 = \frac{3}{2}$ and $c_2 = \frac{1}{2}$

The particular solution is

$$y = \frac{3}{2}e^{3x} + \frac{1}{2}e^{-x} - 2e^x - 2xe^{-x}$$

b) Find solution, satisfying $y(0) = 0, y'(0) = 0$

a)

Consider $y'' - 2y' -3y = 0$

Assume $y = e^{rx}$, then the characteristic polynomial of the equation is given by $r^2 -2r -3 = 0$

This simplifies to $(r - 3)(r +1) = 0$ which gives us two roots $r_1 = 3$ and $r_2 = -1$

Then the complementary solution is $y_c = c_1e^{3x} + c_2e^{-x}$

Since $y'' - 2y' -3y = 16cosh = 8e^x + 8e^{-x}$

We guess the particular solution is of the form

$y = Ae^x + Bxe^{-x}$

$y' = Ae^x + B(e^{-x} - xe^{-x}) = Ae^x + Be^{-x} - Bxe^{-x}$

$y'' = Ae^x - Be^{-x} - B(e^{-x} - xe^{-x}) = Ae^x - 2Be^{-x} + Bxe^{-x}$

Substituting back to the original equation:

$y'' - 2y' -3y = Ae^x - 2Be^{-x} + Bxe^{-x} - 2Ae^x - 2Be^{-x} + 2Bxe^{-x} - 3Ae^x - 3Bxe^{-x} = -4Ae^x - 4Be^{-x} = 8e^x + 8e^{-x}$

Therefore, $A = -2$ and $B = -2$

The particular solution is $y_p = - 2e^x - 2xe^{-x}$

The general solution is :

$$y = c_1e^{3x} + c_2e^{-x} - 2e^x - 2xe^{-x}$$

b)

$$y = c_1e^{3x} + c_2e^{-x} - 2e^x - 2xe^{-x}$$

$$y' = 3c_1e^{3x} - c_2e^{-x} - 2e^x + 2xe^{-x} - 2e^{-x}$$

Plug in the initial conditions $y(0) = 0, y'(0) = 0$, we have,

$0 = c_1 + c_2 - 2$

$0 = 3c_1 - c_2 - 2 - 2$

so, $c_1 = \frac{3}{2}$ and $c_2 = \frac{1}{2}$

The particular solution is

$$y = \frac{3}{2}e^{3x} + \frac{1}{2}e^{-x} - 2e^x - 2xe^{-x}$$

6

Here is another method to find $\varphi$. (By integrating N with respect to y)

Find integrating factor and then a general solution of the ODE

$$(y+3y^2e^{2x}) + (1+2ye^{2x})y' = 0, y(0) = 1$$

Let $M = y+3y^2e^{2x}$ and $N = 1+2ye^{2x}$

We can see that $M_y = 1 + 6e^{2x}y$ and $N_x = 4e^{2x}y$

They are not equal, so not exact. Our goal is to find an integrating factor and make $M_y$ equals $N_x$

$$R_2 = \frac{M_y - N_x}{N} = \frac{1 + 2e^{2x}y}{1+2e^{3x}y} = 1$$

So $\mu = e^{\int R_2dx} = e^{\int1dx} = e^x$

Multiply $\mu$ on both side of the orignial equation and we got

$$(e^xy+3y^2e^{3x}) + (e^x+2ye^{3x})y' = 0$$

Now $M_y = e^x + 6e^{3x}y$ and $N_x = e^x + 6e^{3x}y$

Therefore $\exists\varphi_{(x,y)}$ satisfy $\varphi_x = M$ and $\varphi_y = N$

$$\varphi = \int Ndy = \int (e^x+2ye^{3x})dy = e^xy+y^2e^{3x} + h(x)$$

Then $\varphi_x =e^xy +3y^2e^{3x} + h'(x)$

Since $\varphi_x = M = e^xy+3y^2e^{3x}$

So $h'(x) =0$ and $h(x)=c$

$$\varphi = e^xy+y^2e^{3x} = c$$

Given initial condition $y(0) = 1$

We have $1\times 1 + 1^2 \times 1 = c$ and $c = 2$

Therefore, the particular solution is

**$$e^xy+y^2e^{3x} = 2$$**

Find integrating factor and then a general solution of the ODE

$$(y+3y^2e^{2x}) + (1+2ye^{2x})y' = 0, y(0) = 1$$

Let $M = y+3y^2e^{2x}$ and $N = 1+2ye^{2x}$

We can see that $M_y = 1 + 6e^{2x}y$ and $N_x = 4e^{2x}y$

They are not equal, so not exact. Our goal is to find an integrating factor and make $M_y$ equals $N_x$

$$R_2 = \frac{M_y - N_x}{N} = \frac{1 + 2e^{2x}y}{1+2e^{3x}y} = 1$$

So $\mu = e^{\int R_2dx} = e^{\int1dx} = e^x$

Multiply $\mu$ on both side of the orignial equation and we got

$$(e^xy+3y^2e^{3x}) + (e^x+2ye^{3x})y' = 0$$

Now $M_y = e^x + 6e^{3x}y$ and $N_x = e^x + 6e^{3x}y$

Therefore $\exists\varphi_{(x,y)}$ satisfy $\varphi_x = M$ and $\varphi_y = N$

$$\varphi = \int Ndy = \int (e^x+2ye^{3x})dy = e^xy+y^2e^{3x} + h(x)$$

Then $\varphi_x =e^xy +3y^2e^{3x} + h'(x)$

Since $\varphi_x = M = e^xy+3y^2e^{3x}$

So $h'(x) =0$ and $h(x)=c$

$$\varphi = e^xy+y^2e^{3x} = c$$

Given initial condition $y(0) = 1$

We have $1\times 1 + 1^2 \times 1 = c$ and $c = 2$

Therefore, the particular solution is

7

a) Find Wronskian $W(y_1, y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE $x^2y'' -2xy' + (x^2+2)y = 0$

b) Check that $y_1(x) = xcosx$ is a solution and find another linearly independent solution.

c) Write the general solution, and find solution that $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$

a)

$$y'' - \frac{2}{x}y' + (1 + \frac{2}{x^2}) = 0$$

We see that $p(x) = -\frac{2}{x}$ is continuous everywhere except at $x=0$, $q(x) = (1 + \frac{2}{x^2})$ is continuous everywhere except at $x=0$.

Then by Abel's Theorem,

$$W(y_1, y_2)(x) = ce^{-\int p(x)dx} = ce^{\int(\frac{2}{x})dx} = ce^{2lnx} = cx^2$$

b)

Let's verify $y_1(x) = xcosx$ is a valid solution.

Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$

Plug in: $x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$

So $y_1(x) = xcosx$ is a solution.

Take $c = 1, W(y_1, y_2)(x) = x^2$.

By Reduction of Order, we have:

$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$

c)

Then by part b, the general solution is $y = c_1xcosx +c_2xsinx$

The derivative is given by $y' = c_1(cosx -xsinx) + c_2(xcosx + sinx)$

Plug in the initial conditions $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$, we have

$1 = \frac{\pi}{2}c_2$ and $0 = (-\frac{\pi}{2})c_1 + c_2$

So $c_2 = \frac{2}{\pi}$ and $c_1 = \frac{4}{(\pi)^2}$

The particular solution is,

$$y = \frac{4}{(\pi)^2}xcosx +\frac{2}{\pi}xsinx$$

b) Check that $y_1(x) = xcosx$ is a solution and find another linearly independent solution.

c) Write the general solution, and find solution that $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$

a)

$$y'' - \frac{2}{x}y' + (1 + \frac{2}{x^2}) = 0$$

We see that $p(x) = -\frac{2}{x}$ is continuous everywhere except at $x=0$, $q(x) = (1 + \frac{2}{x^2})$ is continuous everywhere except at $x=0$.

Then by Abel's Theorem,

$$W(y_1, y_2)(x) = ce^{-\int p(x)dx} = ce^{\int(\frac{2}{x})dx} = ce^{2lnx} = cx^2$$

b)

Let's verify $y_1(x) = xcosx$ is a valid solution.

Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$

Plug in: $x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$

So $y_1(x) = xcosx$ is a solution.

Take $c = 1, W(y_1, y_2)(x) = x^2$.

By Reduction of Order, we have:

$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$

c)

Then by part b, the general solution is $y = c_1xcosx +c_2xsinx$

The derivative is given by $y' = c_1(cosx -xsinx) + c_2(xcosx + sinx)$

Plug in the initial conditions $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$, we have

$1 = \frac{\pi}{2}c_2$ and $0 = (-\frac{\pi}{2})c_1 + c_2$

So $c_2 = \frac{2}{\pi}$ and $c_1 = \frac{4}{(\pi)^2}$

The particular solution is,

$$y = \frac{4}{(\pi)^2}xcosx +\frac{2}{\pi}xsinx$$

8

b) Forgot to check $y_1(x) = xcosx$ is a valid solution.

Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$

Plug in: $x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$

So $y_1(x) = xcosx$ is a solution.

Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$

Plug in: $x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$

So $y_1(x) = xcosx$ is a solution.

9

Could also use Reduction of Order to solve part b).

Take $c = 1, W(y_1, y_2)(x) = x^2$.

By Reduction of Order, we have:

$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$

Take $c = 1, W(y_1, y_2)(x) = x^2$.

By Reduction of Order, we have:

$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$

10

Find the general solution of the given differential equation:

$$9y'' + 6y' + y = 0$$

Solve:

The characteristic polynomial is given by $9r^2 + 6r + 1 = 0$

Factor it we obtain $(3r + 1)(3r + 1)$

Then $r_1 = -\frac{1}{3}$ and $r_2 = -\frac{1}{3}$

We observe repeated roots here.

Then the general solution of the given differential equation is $y(t) = c_1e^{-\frac{1}{3} t} + c_2te^{-\frac{1}{3}t}$

$$9y'' + 6y' + y = 0$$

Solve:

The characteristic polynomial is given by $9r^2 + 6r + 1 = 0$

Factor it we obtain $(3r + 1)(3r + 1)$

Then $r_1 = -\frac{1}{3}$ and $r_2 = -\frac{1}{3}$

We observe repeated roots here.

Then the general solution of the given differential equation is $y(t) = c_1e^{-\frac{1}{3} t} + c_2te^{-\frac{1}{3}t}$

11

Find the solution of the given initial problem:

$$y''+4y'+3y = 0, y(0) = 2, y'(0) = -1$$

Assume $y = e^{rt}$, then it must follow that r is the root of the characteristic polynomial

$$r^2+4r+3=0\\

(r+1)(r+3)=0$$

We have $r_1 = -1$ or $r_2 = -3$.

The general solution of the second order differential equation has the form of

$$y = c_1e^{r_1t} + c_2e^{r_2t}$$

Thus, we have

$$y = c_1e^{-t} + c_2e^{-3t}$$

The derivative of this general solution is

$$y' = -c_1e^{-t} - 3c_2e^{-3t}$$

To satisfy both initial conditions $y(0) = 2$ and $y'(0) = -1$,

We have $2 = c_1 + c_2$ and $-1 = -c_1 -3c_2$

This gives us $c_1 = \frac{5}{2}$ and $c_2 = -\frac{1}{2}$

Therefore, the solution of the initial value problem is

$$y = \frac{5}{2}e^{-t} -\frac{1}{2}e^{-3t}$$

$$y''+4y'+3y = 0, y(0) = 2, y'(0) = -1$$

Assume $y = e^{rt}$, then it must follow that r is the root of the characteristic polynomial

$$r^2+4r+3=0\\

(r+1)(r+3)=0$$

We have $r_1 = -1$ or $r_2 = -3$.

The general solution of the second order differential equation has the form of

$$y = c_1e^{r_1t} + c_2e^{r_2t}$$

Thus, we have

$$y = c_1e^{-t} + c_2e^{-3t}$$

The derivative of this general solution is

$$y' = -c_1e^{-t} - 3c_2e^{-3t}$$

To satisfy both initial conditions $y(0) = 2$ and $y'(0) = -1$,

We have $2 = c_1 + c_2$ and $-1 = -c_1 -3c_2$

This gives us $c_1 = \frac{5}{2}$ and $c_2 = -\frac{1}{2}$

Therefore, the solution of the initial value problem is

$$y = \frac{5}{2}e^{-t} -\frac{1}{2}e^{-3t}$$

12

From what I understand, when you are solving the second order initial value problems, you first get a general solution of that equation with two arbitrary constant of the form y(t)=C_{1}e^{rt}+C_{2}e^{rt}. Then you find the derivative y' of this general solution, and plug in the initial conditions to solve what the two constants are. So you might not necessary need both initial conditions to be defined at t_{0}.

13

Determine whether the equation given below is exact. If it is exact, find the solution

$$

(e^xsin(y)-2ysin(x))-(3x-e^xsin(y))y'=0\\

(e^xsin(y)-2ysin(x))dx-(3x-e^xsin(y))dy=0

$$

Let $M(x,y) = e^xsin(y)-2ysin(x)$

Let $N(x,y) = -(3x-e^xsin(y)) = e^xsin(y)-3x$

$$

M_y(x,y) = \frac{\partial}{\partial y} M(x,y)= \frac{\partial}{\partial y} (e^xsin(y)-2ysin(x)) = e^xcos(y)-2sin(x)\\

N_x(x,y) = \frac{\partial}{\partial x} N(x,y) = \frac{\partial}{\partial x} (e^xsin(y)-3x) = e^xsin(y)-3

$$

Clearly, we see that $ e^xcos(y)-2sin(x) \neq e^xsin(y)-3$

Therefore, $M_y(x,y) \neq N_x(x,y)$

By definition of exact, we can conclude the equation is not exact.

$$

(e^xsin(y)-2ysin(x))-(3x-e^xsin(y))y'=0\\

(e^xsin(y)-2ysin(x))dx-(3x-e^xsin(y))dy=0

$$

Let $M(x,y) = e^xsin(y)-2ysin(x)$

Let $N(x,y) = -(3x-e^xsin(y)) = e^xsin(y)-3x$

$$

M_y(x,y) = \frac{\partial}{\partial y} M(x,y)= \frac{\partial}{\partial y} (e^xsin(y)-2ysin(x)) = e^xcos(y)-2sin(x)\\

N_x(x,y) = \frac{\partial}{\partial x} N(x,y) = \frac{\partial}{\partial x} (e^xsin(y)-3x) = e^xsin(y)-3

$$

Clearly, we see that $ e^xcos(y)-2sin(x) \neq e^xsin(y)-3$

Therefore, $M_y(x,y) \neq N_x(x,y)$

By definition of exact, we can conclude the equation is not exact.

15

a very small typo: the final solution should be just y, not y^2.

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