16

**MAT244--Lectures & Home Assignments / Sec 7.1 Q4**

« **on:**October 30, 2018, 12:05:54 PM »

What does $u^{(4)}$ stand for?

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

Pages: 1 [**2**]

16

What does $u^{(4)}$ stand for?

17

(a) reformat equation into

$$y'' + p(x)y' + q(x)$$

where

$$p(x) = -\frac{x}{x-1}$$

by equation of wronskian

\begin{gather*}

W(x) = \exp(-\int p(x) dx)\\

W(x) = \exp(\int \frac{x}{x-1} dx) = \exp(\int \frac{u+1}{u}du) = c_0ue^u = c_0(xe^{x-1} - e^{x-1})

\end{gather*}

choose $c_0 = e$

$$W(x) = xe^x - e^x$$

(b) Plug in $y = x, y' = 1, y'' = 0$, equation becomes

$$-x + x = 0$$

therefore $y=x$ is a solution, let second solution be $y_2$ and let $y_1 = x$ since

$$W(x) = y_1y_2' - y_1'y_2$$

Plug in $y_1 = x, y_1' = 1$

$$xe^x - e^x = xy_2' - y_2$$

By inspection

$$y_2 = e^x$$

Hence, general solution is

$$y(x) = c_1x + c_2e^x$$

c) Set $y(0) = 1$

$$c_2 = 1$$

Set $y'(0) = 0, y'(0) = c_1 + c_2e^x$

$$c_1 + c_2 = 0, c_1 = -c_2 = -1$$

Hence the solution is

$$y(x) = -x + e^x$$

$$y'' + p(x)y' + q(x)$$

where

$$p(x) = -\frac{x}{x-1}$$

by equation of wronskian

\begin{gather*}

W(x) = \exp(-\int p(x) dx)\\

W(x) = \exp(\int \frac{x}{x-1} dx) = \exp(\int \frac{u+1}{u}du) = c_0ue^u = c_0(xe^{x-1} - e^{x-1})

\end{gather*}

choose $c_0 = e$

$$W(x) = xe^x - e^x$$

(b) Plug in $y = x, y' = 1, y'' = 0$, equation becomes

$$-x + x = 0$$

therefore $y=x$ is a solution, let second solution be $y_2$ and let $y_1 = x$ since

$$W(x) = y_1y_2' - y_1'y_2$$

Plug in $y_1 = x, y_1' = 1$

$$xe^x - e^x = xy_2' - y_2$$

By inspection

$$y_2 = e^x$$

Hence, general solution is

$$y(x) = c_1x + c_2e^x$$

c) Set $y(0) = 1$

$$c_2 = 1$$

Set $y'(0) = 0, y'(0) = c_1 + c_2e^x$

$$c_1 + c_2 = 0, c_1 = -c_2 = -1$$

Hence the solution is

$$y(x) = -x + e^x$$

18

Also, there is an earlier one in Sec2.4 27, where $y'$ should be replaced by $z'$. Not sure if a separate post is appropriate.

19

In **Method of Reduction**, part a, after wronskian is expanded $y_1$ should be replaced by $y_1'$

20

Reformat equation into form $y'' + p(t)y' + q(t) = 0$

We can see that

$p(t) = \frac{\sin(t)}{\cos(t)}$

By the equation for wronskian

$$W(t) = \exp \bigl(\int p(t) dt\bigr) = \exp \bigl(\int \frac{\sin(t)}{\cos(t)} dt\bigr) = \exp (-\ln(\cos(t))+\ln (c_0)) = \frac{c_0}{\cos(t)}$$

where $c_0$ depends on choice of $y_2$ and $y_2$

We can see that

$p(t) = \frac{\sin(t)}{\cos(t)}$

By the equation for wronskian

$$W(t) = \exp \bigl(\int p(t) dt\bigr) = \exp \bigl(\int \frac{\sin(t)}{\cos(t)} dt\bigr) = \exp (-\ln(\cos(t))+\ln (c_0)) = \frac{c_0}{\cos(t)}$$

where $c_0$ depends on choice of $y_2$ and $y_2$

21

$y_1 = x^3, y_1' = 3x^2, y_1'' = 6x$

$y_2 = y_2' = y_2'' = e^x$

Expand W(y, y_1, y_2)

$$y(y_1'y_2'' - y_2'y_1'') - y'(y_1y_2'' - y_2y_1'') + y''(y_1y_2' - y_2y_1') = 0.$$

Plug in $y_1$, $y_2$ and their derivatives, equation becomes

$$(x^3 - 3x^2)e^xy'' + (6x - x^3)e^xy' + (3x^2 - 6x)e^xy' = 0$$

or

$$ (x^2 - 3x)y'' + (6 - x^2)y' + (3x - 6)y' = 0$$

Btw, I think typo is still there. Two $y_2''$. FIXED

$y_2 = y_2' = y_2'' = e^x$

Expand W(y, y_1, y_2)

$$y(y_1'y_2'' - y_2'y_1'') - y'(y_1y_2'' - y_2y_1'') + y''(y_1y_2' - y_2y_1') = 0.$$

Plug in $y_1$, $y_2$ and their derivatives, equation becomes

$$(x^3 - 3x^2)e^xy'' + (6x - x^3)e^xy' + (3x^2 - 6x)e^xy' = 0$$

or

$$ (x^2 - 3x)y'' + (6 - x^2)y' + (3x - 6)y' = 0$$

Btw, I think typo is still there. Two $y_2''$. FIXED

22

\begin{gather*}

N_x = \frac{2x}{y} - \frac{3y}{x^2}, M_y = -\frac{6}{y^2}, xM = 3x^2 + \frac{6x}{y}, yN = x^2 + 3\frac{y^2}{x}, \\\frac{N_x - M_y}{xM - yN} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{2x^2 - \frac{3y^2}{x} + \frac{6x}{y}} = R = \frac{1}{xy}

\end{gather*}

From assignment

\begin{gather*}

\mu (z) = \exp(\int \frac{1}{z} d(z)) = z,\\

\mu M = 3x^2y + 6x, \mu N = x^3 + 3y^2,\\

\int \mu M dx = x^3y + 3x^2 + f(y) + c_0,\\

\int \mu N dy = x^3y + y^3 + g(x) + c_1.

\end{gather*}

Combine the previous two result gives

$$ \phi(x, y) = x^3y + 3x^2 + y^3 = c,$$

~~where ~~

$\frac{\partial\phi}{\partial x} = uM, \frac{\partial\phi}{\partial x} = uN$ We do not need this

N_x = \frac{2x}{y} - \frac{3y}{x^2}, M_y = -\frac{6}{y^2}, xM = 3x^2 + \frac{6x}{y}, yN = x^2 + 3\frac{y^2}{x}, \\\frac{N_x - M_y}{xM - yN} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{2x^2 - \frac{3y^2}{x} + \frac{6x}{y}} = R = \frac{1}{xy}

\end{gather*}

From assignment

\begin{gather*}

\mu (z) = \exp(\int \frac{1}{z} d(z)) = z,\\

\mu M = 3x^2y + 6x, \mu N = x^3 + 3y^2,\\

\int \mu M dx = x^3y + 3x^2 + f(y) + c_0,\\

\int \mu N dy = x^3y + y^3 + g(x) + c_1.

\end{gather*}

Combine the previous two result gives

$$ \phi(x, y) = x^3y + 3x^2 + y^3 = c,$$

$\frac{\partial\phi}{\partial x} = uM, \frac{\partial\phi}{\partial x} = uN$

23

I believe there is a typo in exercise 2.4 question 27b.

I was checking the textbook and seems like the substitution should be $v = y^{1-n}$ though what I see from course website is $v = y^{1\hat{a}'n}$

I was checking the textbook and seems like the substitution should be $v = y^{1-n}$ though what I see from course website is $v = y^{1\hat{a}'n}$

24

Rephrase equation

$y' + \frac{2}{t}y = \frac{sin(t)}{t}$

Find integrating factor

$u(t) = e^{\int \frac{2}{t}} = t^2$

the constant from integration is chosen to be zero. Now

$y = \frac{1}{u(t)}\int u(t)\frac{sin(t)}{t}$

Use int by parts,

$y = -\frac{cos(t)}{t} + \frac{sin(t)}{t^2} + \frac{c_1}{t^2}$

Since $t$ is in denominator and the nominators are bounded, as $t \to \infty$, $y \to 0$

$y' + \frac{2}{t}y = \frac{sin(t)}{t}$

Find integrating factor

$u(t) = e^{\int \frac{2}{t}} = t^2$

the constant from integration is chosen to be zero. Now

$y = \frac{1}{u(t)}\int u(t)\frac{sin(t)}{t}$

Use int by parts,

$y = -\frac{cos(t)}{t} + \frac{sin(t)}{t^2} + \frac{c_1}{t^2}$

Since $t$ is in denominator and the nominators are bounded, as $t \to \infty$, $y \to 0$

25

First divide by $t^3$ on both side of the equation, we get

$$y' + \frac{4}{t}y = \frac{e^{-t}}{t^3}$$

Using the method of integrating factor we have equation for $u(t)$

$$u(t) = e^{\int \frac{4}{t}dt} = e^{4\ln(t) + c} = t^4$$

where constant $c$ is arbitrary, it's chosen to be 0 here. Then

$$\bigl(y u(t)\bigr)' = u(t)\frac{e^{-t}}{t^3}$$

rearranging gives equation

$$y = \frac{1}{u(t)}\int u(t)\frac{e^{-t}}{t^3}$$

substitute in $u(t) = t^4$

$$y = \frac{1}{t^4}\int te^{-t}$$

use integration by parts

$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} + \frac{c_1}{t^4}$$

to check $c_1$, plug in condition $y(-1) = 0$

$$y(-1) = e - e + c_1 = c_1= 0$$

Plug in $c_1 = 0$ gets

$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} $$

$$y' + \frac{4}{t}y = \frac{e^{-t}}{t^3}$$

Using the method of integrating factor we have equation for $u(t)$

$$u(t) = e^{\int \frac{4}{t}dt} = e^{4\ln(t) + c} = t^4$$

where constant $c$ is arbitrary, it's chosen to be 0 here. Then

$$\bigl(y u(t)\bigr)' = u(t)\frac{e^{-t}}{t^3}$$

rearranging gives equation

$$y = \frac{1}{u(t)}\int u(t)\frac{e^{-t}}{t^3}$$

substitute in $u(t) = t^4$

$$y = \frac{1}{t^4}\int te^{-t}$$

use integration by parts

$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} + \frac{c_1}{t^4}$$

to check $c_1$, plug in condition $y(-1) = 0$

$$y(-1) = e - e + c_1 = c_1= 0$$

Plug in $c_1 = 0$ gets

$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} $$

26

Check the arrow on quiz 1. http://www.math.toronto.edu/courses/mat244h1/20189/homeassignments-fall.html

27

Maybe in that website independent variable is $t$ and $x$, $y$ are the dependent variables. That could be why it's said to be nonlinear.

28

Will the quizzes cover materials based on lectures, tutorials, or assignments due before the date of the quiz?

29

When your equation for $y'$ only has dependence on $y$, directional field has same value for same $y$.

But $y'$ could have dependence on other variables like $t$. $y' = y + t$ for example, even with $y$ fixed, could yield different slope at different $t$.

But $y'$ could have dependence on other variables like $t$. $y' = y + t$ for example, even with $y$ fixed, could yield different slope at different $t$.

30

dt can exist alone. Look up the technique called separation of variable.

Pages: 1 [**2**]