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### Messages - Wusijia

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##### Quiz-4 / TUT0502 QUIZ4
« on: October 18, 2019, 02:01:53 PM »
\begin{equation*}
9y''+6y'+y=0
\end{equation*}
\begin{equation*}
9r^2+6r+1=0
\end{equation*}
\begin{equation*}
(3r+1)^2=0
\end{equation*}
\begin{equation*}
r_1=r_2=-\frac{1}{3}\\
\end{equation*}
\begin{equation*}
y=C_1e^{-\frac{1}{3}t}+C_2te^{\frac{1}{3}t}
\end{equation*}

2
##### Quiz-3 / TUT0502 QUIZ3
« on: October 11, 2019, 02:00:02 PM »
\begin{equation*}
y''-2y'-2y=0
\end{equation*}
\begin{equation*}
r^2-2r-2=0
\end{equation*}
\begin{equation*}
r_1=\frac{2+\sqrt{4+8}}{2}=1+\sqrt{3}
\end{equation*}
\begin{equation*}
r_2=\frac{2-\sqrt{4+8}}{2}=1-\sqrt{3}
\end{equation*}
\begin{equation*}
y=C_1e^{(1+\sqrt{3})t}+C_2e^{(1-\sqrt{3})t)}
\end{equation*}

3
##### Quiz-2 / TUT0502 QUIZ2
« on: October 05, 2019, 07:26:32 PM »
\begin{equation*}
((x+2)sin(y))+xcos(y)y'=0
\end{equation*}
\begin{equation*}
\mu=xe^x
\end{equation*}
\begin{equation*}
M_y=(x+2)cosy
\end{equation*}
\begin{equation*}
N_x=cosy
\end{equation*}
\begin{equation*}
M_y\neq N_x \Rightarrow not \ exact
\end{equation*}
\begin{equation*}
multiple \ \mu \ on \ both \ sides
\end{equation*}
\begin{equation*}
xe^x((x+2)sin(y))+x^2e^xcos(y)y'=0
\end{equation*}
\begin{equation*}
(x^2e^x+2xe^x)sin(y)+x^2e^xcos(y)y'=0
\end{equation*}
\begin{equation*}
M_y=(x^2e^x+2xe^x)cosy
\end{equation*}
\begin{equation*}
N_x=cosy(2xe^x+x^2e^x)
\end{equation*}
\begin{equation*}
M_y=N_x\Rightarrow exact
\end{equation*}
\begin{equation*}
\varphi_y=N=x^2e^xcosy
\end{equation*}
\begin{equation*}
\varphi=x^2e^xsiny+h(y)
\end{equation*}
\begin{equation*}
\varphi_x=M \Rightarrow (x^2e^x+2xe^x)siny=2xe^xsiny+x^2e^xsiny+h'(y)
\end{equation*}
\begin{equation*}
x^2e^xsiny+2xe^xsiny=2xe^xsiny+x^2e^xsiny+h'(y)
\end{equation*}
\begin{equation*}
h'(y)=0
\end{equation*}
\begin{equation*}
h(0)=0
\end{equation*}
\begin{equation*}
\varphi=x^2e^xsiny+0=x^2e^xsiny
\end{equation*}
\begin{equation*}
x^2e^xsiny=C
\end{equation*}

4
##### Quiz-1 / TUT0502 QUIZ1
« on: September 27, 2019, 03:08:59 PM »
\begin{equation*}
ty'+(t+1)y=t\ y(ln2)=1\ t>0
\end{equation*}
\begin{equation*}
y'+ \frac{t+1}{t}y=1
\end{equation*}
\begin{equation*}
\mu=e^{\int\frac{t+1}{t}dt}=e^{\int1+\frac{1}{t}dt}=e^{lnt+t}=te^t
\end{equation*}
\begin{equation*}
te^ty'+e^t(t+1)y=te^t
\end{equation*}
\begin{equation*}
(te^ty)'=te^t
\end{equation*}
\begin{equation*}
\int (te^ty)'dt=\int te^tdt
\end{equation*}
\begin{equation*}
{Let} \ u=t \ du=dt \ dv=e^tdt\  v=e^t
\end{equation*}
\begin{equation*}
\int te^tdt=te^t-\int e^tdt=te^t-e^t+C
\end{equation*}
\begin{equation*}
te^ty=te^t-e^t+C
\end{equation*}
\begin{equation*}
y=1-\frac{1}{t}+\frac{C}{te^t}
\end{equation*}
\begin{equation*}
{Since} \ y(ln2)=1
\end{equation*}
\begin{equation*}
1=1-\frac{1}{ln2}+\frac{C}{(ln2)e^{ln2}}
\end{equation*}
\begin{equation*}
\frac{1}{ln2}=\frac{C}{2ln2}
\end{equation*}
\begin{equation*}
C=2
\end{equation*}
\begin{equation*}
{Therefore,}y=1-\frac{1}{t}+\frac{2}{te^t} {,where} \ t>0
\end{equation*}

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