# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:24:23 AM

Title: TT1 Problem 2 (main)
Post by: Victor Ivrii on October 16, 2018, 05:24:23 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*}
(x-1)y''-xy'+y=0.
\end{equation*}

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

(c) Write the general solution, and find solution such that ${y(0)=1, y'(0)=0}$.
Title: Re: TT1 Problem 2 (main)
Post by: Tzu-Ching Yen on October 16, 2018, 07:29:31 AM
(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
$$y(x) = -x + e^x$$
Title: Re: TT1 Problem 2 (main)
Post by: Victor Ivrii on October 16, 2018, 07:31:06 AM
Thomson, please clean up. Also never use * for multiplication in any math class.

Now it is fine!