Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 1 => Topic started 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^xe^{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 ${y(0)=0, y'(0)=2}$.

(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
$$\begin{align}e^{x}(e^{x}+e^{x}+2)+e^{x}(e^{x}e^{x})+2(e^{x}+1)&=0\\2e^xe^{2x}2+2e^{x}+e^{2x}+2&=0\end{align}$$
$y_1(x)$ 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
$$y(x)=c_1 (e^x+1)+c_2(1\frac{1}{e^{x}}).$$
$\Rightarrow y'(x)=c_1e^x+c_2e^{x}$
$$\cases{c_1c_2=0\\c_1+c_2=2} \Rightarrow \cases{c_1=1\\c_2=1}$$
Thus, the solution that satisfies $y(0)=1$, $y'(0)=2$ is
$$y(x)=e^x\frac{1}{e^{x}}.$$

Hi, there is my solution for Problem 2

$y''  \frac{e^xe^{x}}{e^x+e^{x}+2} y'  \frac{2}{e^x+e^{x}+2}y = 0$
$W = Ce^{\int{\frac{e^xe^{x}}{e^x+e^{x}+2}}dx} = C{e^x+e^{x}+2} $
$take C =1, W = e^x+e^{x}+2$

Hi, there is my solution for Problem 2
Hey Doris, you wrote the definition of the Wronskian incorrectly, $W(y_1,y_2)(x)\ne y_1(x)y_2’(x)+y_2(x)y_1'(x)$.
Instead,
$$W(y_1,y_2)(x)=\begin{array}{c c}y_1(x)& y_2(x) \\ y_1’(x) & y_2’(x)\end{array}=y_1(x)y_2’(x)y_2(x)y_1'(x).$$
Hope that helps!

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

Integration of (cosh(x) + 1)/(e^x + 1)^2 wrt dx = e^(x)/2 + c
Therefore g = (e^x + 1)(e^(x)/2 + c)

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.