# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:28:13 AM

Title: TT1 Problem 2 (morning)
Post by: Victor Ivrii on October 16, 2018, 05:28:13 AM
(a)  Find Wronskian  $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x) , y_2(x)$ for ODE
\begin{equation*}
-y'' (e^x+e^{-x}+2) + y' (e^x-e^{-x}) +2y=0.
\end{equation*}
(b) Check that $y_1(x)=e^x+1$ is a solution and find another linearly independent solution.

(c) Write the general solution, and find solution such that ${y(0)=0, y'(0)=2}$.
Title: Re: TT1 Problem 2 (morning)
Post by: Monika Dydynski on October 16, 2018, 11:23:09 AM
(a) Find Wronskian $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE

$$-y''(e^{x}+e^{-x}+2)+y'(e^{x}-e^{-x})+2y=0.$$

Dividing but sides by $-(e^{x}+e^{-x}+2)$, we get

$$L[y]=y''-y'\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}-y\frac{2}{e^{x}+e^{-x}+2}=0,$$

where $p(x)=-\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}$, and $q(t)=-\frac{2}{e^{x}+e^{-x}+2}$.

By Abel's Theorem,

\begin{align}W(y_1,y_2)(x)&=c\exp(\int-{p(x)dx})\\&=c\exp(\int\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}dx)\\&=c (e^x+e^{-x}+2).\end{align}

Let $c=1 \Rightarrow W(y_1,y_2)(x)=e^x+e^{-x}+2$.

b) Check that $y_1(x)=e^{x}+1$ is a solution and find another linearly independent solution.

Since $y_1(x)=e^{x}+1 \Rightarrow y_1 '(x)=e^{x}$, and $y_1 ''(x)=e^{x}$

Plugging $y_1$, $y_1 '$, and $y_1 ''$ into the ODE, we have

\begin{align}-e^{x}(e^{x}+e^{-x}+2)+e^{x}(e^{x}-e^{-x})+2(e^{x}+1)&=0\\-2e^x-e^{2x}-2+2e^{x}+e^{2x}+2&=0\end{align}

$y_1(x)$ satisfies the ODE $\Rightarrow$ $y_1(x)$ is a solution.

Given $y_1(x)$, we can find another linearly independent solution.

We know from the definition of the Wronskian that
$$W(y_1,y_2)(x)=y_1y_2 '-y_1 'y_2=(e^x+1)y_2 '-y_2 e^x$$

Equating the two expressions for the Wronskian, we get

$$(e^x+1)y_2 '-y_2 e^x=e^x+e^{-x}+2$$

Dividing both sides by $e^x+1$, and multiplying by integrating factor $\mu=\frac{1}{e^x+1}$,

$$\frac{1}{e^x+1}y_2=\int{\frac{e^x+e^{-x}+2}{e^{2x}+2e^x+1}}dx+C$$
$$\frac{1}{e^x+1}y_2=-\frac{1}{e^x}$$
$$y_2=-\frac{e^x+1}{e^x}$$
$$y_2=-1-\frac{1}{e^{x}}$$

c) Write the general solution. Find solution such that $y(0)=0$, $y'(0)=2$

The general solution to the ODE is

$$y(x)=c_1 (e^x+1)+c_2(-1-\frac{1}{e^{x}}).$$

$\Rightarrow y'(x)=c_1e^x+c_2e^{-x}$

$$\cases{c_1-c_2=0\\c_1+c_2=2} \Rightarrow \cases{c_1=1\\c_2=1}$$

Thus, the solution that satisfies $y(0)=1$, $y'(0)=2$ is

$$y(x)=e^x-\frac{1}{e^{x}}.$$

Title: Re: TT1 Problem 2 (morning)
Post by: Doris Zhuomin Jia on October 16, 2018, 11:27:14 AM
Hi, there is my solution for Problem 2
Title: Re: TT1 Problem 2 (morning)
Post by: Mengmeng Shang on October 16, 2018, 11:32:56 AM
$y'' - \frac{e^x-e^{-x}}{e^x+e^{-x}+2} y' - \frac{2}{e^x+e^{-x}+2}y = 0$

$W = Ce^{-\int{-\frac{e^x-e^{-x}}{e^x+e^{-x}+2}}dx} = C{e^x+e^{-x}+2}$

$take C =1, W = e^x+e^{-x}+2$
Title: Re: TT1 Problem 2 (morning)
Post by: Monika Dydynski on October 16, 2018, 11:44:27 AM
Hi, there is my solution for Problem 2

Hey Doris, you wrote the definition of the Wronskian incorrectly, $W(y_1,y_2)(x)\ne y_1(x)y_2’(x)+y_2(x)y_1'(x)$.

$$W(y_1,y_2)(x)=\begin{array}{|c c|}y_1(x)& y_2(x) \\ y_1’(x) & y_2’(x)\end{array}=y_1(x)y_2’(x)-y_2(x)y_1'(x).$$

Hope that helps!

Title: Re: TT1 Problem 2 (morning)
Post by: Weina Zhu on October 16, 2018, 01:34:56 PM
I am trying to go the cosh~sinh way, but cannot continue when finding the second solution, can someone help me out of this?
Thanks in advance. I tried to see classmates' post, but they gave the answer directly;D
Title: Re: TT1 Problem 2 (morning)
Post by: Samarth Agarwal on October 16, 2018, 02:39:54 PM
Integration of (cosh(x) + 1)/(e^x + 1)^2 wrt dx = -e^(-x)/2 + c
Therefore g = (e^x + 1)(-e^(-x)/2 + c)
Title: Re: TT1 Problem 2 (morning)
Post by: Victor Ivrii on October 18, 2018, 04:36:31 AM
Monika did everything right, I also commend her comment to Doris' post.

Weina
You are right that equation is
$$-2 (\cosh(x)+1)y''+2\sinh(x)y'+2y=0$$
and we can rewrite it as
$$(\cosh(x)+1)y''-\sinh(x)y'-y=0$$
and then
$$W=C\exp\Bigl(\int \frac{\sinh(x)}{\cosh(x)+1}\,dx\Bigr) = C\exp\Bigl(\ln (\cosh(x)+1)\Bigr)=C_1(\cosh(x)+1).$$
But it is not another way, just another form.