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### Messages - zhijian ling

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##### Quiz-5 / LEC5201 quiz5
« on: November 01, 2019, 12:12:37 AM »
Quote
Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given inhomogeneous equation.
\begin{gather*}
(1-t)y'' + ty' - y =2(t-1)^2e^{-t}, \qquad 0 <t<1;\\
y_1(t) = e^t, \qquad y_2(t) = t
\end{gather*}
Solution:
first check if $y_1$,$y_2$ are solutions to the homogeneous equation
\begin{equation*}
\begin{cases}
y_1 = e^t\\
y_1'' = e^t\\
y_1'' = e^t\\
\end{cases}
\\(1-t)e^t + te^t- e^t = 0
\end{equation*}
\begin{equation*}
\begin{cases}
y_2 = t\\
y_2'' = 1\\
y_2''= 0\\
\end{cases}
\\t - t = 0
\end{equation*}
Thus, $y_1, y_2$ are solutions to the homogeneous equation.
Then, find the Wronskian
\begin{equation*}
W(y_1,y_2)(t) =
\begin{vmatrix}
e^t&t\\
e^t&1\\
\end{vmatrix}
= e^t - te^t
\end{equation*}
\begin{equation*}
g(x) = \frac{2(t-1)^2e^{-t}}{1-t} = 2(1-t)e^{-t}\\
c_1 = -\int \frac{2t(1-t)e^{-t}}{e^t - te^t}dt = \int-2te^{-2t}dt = (t+\frac{1}{2})e^{-2t}\\
c_2 = \int\frac{2e^t(1-t)e^{-t}}{e^t - te^t}dt = \int2e^{-t}dt = -2e^{-t}
\end{equation*}
Thus, the particular solution is
$$y_p = ((t+\frac{1}{2})e^{-2t})e^t + (-2e^{-t})t = (\frac{1}{2}-t)e^{-t}$$
[/quote]

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##### Quiz-4 / TUT 0502 quiz4
« on: October 18, 2019, 02:03:30 PM »
TUT 0502 quiz4

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##### Quiz-3 / TUT 0502 quiz3
« on: October 11, 2019, 01:59:58 PM »
TUT 0502 quiz3

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##### Quiz-2 / Re: TUT 0502 quiz2
« on: October 07, 2019, 02:00:29 AM »
quiz2

5
##### Quiz-2 / TUT 0502 quiz2
« on: October 07, 2019, 01:54:53 AM »
\documentclass{article}
\renewcommand{\baselinestretch}{1.5}
\usepackage{amsmath}
\begin{document}
\begin{flushleft}
find an integrating factor and solve the given equation.
\begin{flalign*}
&(3x + \frac{6}{y} ) + (\frac{x^2}{y} + 3\frac{y}{x}) \frac{dy}{dx} = 0&\\
\end{flalign*}
simplify the equation and we will get
\begin{flalign*}
&(3x^2y + 6x)dx +(x^3 + 3y^2)dy = 0&\\
&M_y = 3x^2&\\
&N_x =3x^2&\\
&M_y  =  N_x&\\
\end{flalign*}
so it is exact\\
then find $\mu$
\begin{flalign*}
&\mu= \int(3x^2y+6x)dx&\\
&\mu = x^3y +3x^2 + h(y)&\\
&\mu _y= x^3 + h'(y)&\\
&\mu _y= N&\\
&h'(y) = 3y^2&\\
&h(y) = y^3 + c&\\
&\mu = x^3y + 3x^2 + y^3 +c&\\
&x^3y + 3x^2 + y^2 = c&
\end{flalign*}
solution is $x^3y + 3x^2 + y^2 = c$
\end{flushleft}
\end{document}

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