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### Messages - Mengyuan Wang

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1
##### Term Test 1 / Re: Problem 4 (morning)
« on: October 23, 2019, 07:54:27 AM »

\begin{array}{c}{y^{\prime \prime}-6 y^{\prime}+25 y=16 e^{3 x}+102 \sin x} \\ {y=e^{7 x}} \\ {y^{\prime}=r{e}^{rx} } \\ {y^{\prime \prime}=r^{2} e^{r x}}\end{array}

\begin{array}{l}{r^{2}-6 r+25=0} \\ {r=3 \pm 4 i} \\ {y=c_{1} e^{3 x} \cos 4 x+c_{2} e^{3 x} \sin 4 x}\end{array}

let

\begin{array}{l}{y=A e^{3x} } \\ {y^{\prime}=3 A e^{3x} } \\ {y^{\prime \prime}=9 A e^{3x}}\end{array}

\begin{array}{rl}{(9 A-18 A+25 A) e^{3x}} & {=16 e^{3x} } \\ {16 A e^{3x}} & {=16 e^{3x} } \\ {A} & {=1} \\ {y} & {=e^{3x} }\end{array}

\begin{aligned} \text { Let } y &=A \sin (x)+B \cos (x) \\ & y^{\prime}=A \cos (x)-B \sin (x) \\ y^{\prime \prime} &=-A \sin (x)-B \cos (x) \end{aligned}

(-A+6 B+25 A) \sin (x)+(-B-6 A+25 B) \cos (x)=102 \sin (x)

\begin{array}{l}{24 A+6 B=102} \\ {24 B-6 A=0}\end{array}

\begin{array}{c}{4 A+B=17} \\ {4 B-A=10}\end{array}

\begin{array}{l}{A=4} \\ {B=1}\end{array}

y=4 \sin (x)+\cos (x)

\begin{array}{l}{y=\operatorname{ces}^{3 x} \cos (x)+c_{2} e^{3 x} \sin (4 x)+e^{3 t}+4 \sin (x)+\cos (x)} \\ {y^{\prime}=3 c_{1} e^{3 x} \cos (4 x)-4 c_{1} e^{3 x} \sin (4 x)+3 \epsilon_{2} e^{3 x} \sin (x)+c_{2} e^{3 x}(x)+3 e^{3 x}+4 \cos x} \\ {y(0)=y^{\prime}(0)=0 \quad C_{1}=-2 \quad C_{2}=-\frac{1}{4}} \\ {y=-2 e^{3 x} \cos (4 x)-\frac{1}{4} e^{3 x} \sin (4 x)+e^{3 x}+\cos (x)+4 \sin (x)}\end{array}

\end{document}

2
##### Term Test 1 / Re: Problem 4 (afternoon)
« on: October 23, 2019, 07:53:48 AM »

\begin{array}{c}{y^{\prime \prime}-6 y^{\prime}+25 y=16 e^{3 x}+102 \sin x} \\ {y=e^{7 x}} \\ {y^{\prime}=r{e}^{rx} } \\ {y^{\prime \prime}=r^{2} e^{r x}}\end{array}

\begin{array}{l}{r^{2}-6 r+25=0} \\ {r=3 \pm 4 i} \\ {y=c_{1} e^{3 x} \cos 4 x+c_{2} e^{3 x} \sin 4 x}\end{array}

let

\begin{array}{l}{y=A e^{3x} } \\ {y^{\prime}=3 A e^{3x} } \\ {y^{\prime \prime}=9 A e^{3x}}\end{array}

\begin{array}{rl}{(9 A-18 A+25 A) e^{3x}} & {=16 e^{3x} } \\ {16 A e^{3x}} & {=16 e^{3x} } \\ {A} & {=1} \\ {y} & {=e^{3x} }\end{array}

\begin{aligned} \text { Let } y &=A \sin (x)+B \cos (x) \\ & y^{\prime}=A \cos (x)-B \sin (x) \\ y^{\prime \prime} &=-A \sin (x)-B \cos (x) \end{aligned}

(-A+6 B+25 A) \sin (x)+(-B-6 A+25 B) \cos (x)=102 \sin (x)

\begin{array}{l}{24 A+6 B=102} \\ {24 B-6 A=0}\end{array}

\begin{array}{c}{4 A+B=17} \\ {4 B-A=10}\end{array}

\begin{array}{l}{A=4} \\ {B=1}\end{array}

y=4 \sin (x)+\cos (x)

\begin{array}{l}{y=\operatorname{ces}^{3 x} \cos (x)+c_{2} e^{3 x} \sin (4 x)+e^{3 t}+4 \sin (x)+\cos (x)} \\ {y^{\prime}=3 c_{1} e^{3 x} \cos (4 x)-4 c_{1} e^{3 x} \sin (4 x)+3 \epsilon_{2} e^{3 x} \sin (x)+c_{2} e^{3 x}(x)+3 e^{3 x}+4 \cos x} \\ {y(0)=y^{\prime}(0)=0 \quad C_{1}=-2 \quad C_{2}=-\frac{1}{4}} \\ {y=-2 e^{3 x} \cos (4 x)-\frac{1}{4} e^{3 x} \sin (4 x)+e^{3 x}+\cos (x)+4 \sin (x)}\end{array}

\end{document}

3
##### Term Test 1 / Re: Problem 3 (morning)
« on: October 23, 2019, 07:44:36 AM »

\begin{array}{l}{y^{\prime \prime}-6 y^{\prime}+8 y=48 \sinh (2 x)} \\ {y^{\prime \prime}-6 y^{\prime}+8 y=24 e^{2 x}-24 e^{-x}}\end{array}

\text { Let } y=e^{r x}

\begin{array}{l}{y^{\prime}=\sec ^{r x}} \\ {y^{\prime \prime}=r^{2} e^{r x}}\end{array}

\begin{array}{c}{y^{2}-6 x+8=0} \\ {r_{1}=2 \quad r_{2}=4} \\ {y=c_{1} e^{2x} +c_{2} e^{4x} }\end{array}

\text { let } y=A x e^{2 x}

y^{\prime}=A e^{2 x}+2 A x e^{2 x}

y^{\prime \prime}=4 A xx^{2 x}+4 A e^{2 x}

\begin{array}{c}{(8 A-12 A+4A) t \cdot e^{2 x}+(-6 A+4 A) e^{2 x}=24 e^{2 x}} \\ {-2 Ae^{2 x}=24 e^{2 x}} \\ {A=-12} \\ {y=-12 x e^{2 x}}
\end{array}

Let

\begin{array}{l}{y=A e^{-2 x}} \\ {y^{\prime}=-2 A e^{-2 x}} \\ {y^{\prime \prime}=4 A e^{-2 x}}\end{array}

\begin{aligned}(8 A+12 A+4 A) e^{-2 x} &=-24 e^{-2 x} \\ A &=-1 \\ y &=e^{-2 x} \end{aligned}

so

\begin{array}{l}{y=a e^{2}+c_{2} e^{4 x}-12 e^{2 x}-e^{-2 x}} \\ {y^{\prime}=2 c_{1} e^{2 t}+4 c_{2} e^{4 t}-24 r e^{2 x}+2 e^{-2 x}-12 e^{2 x}} \\ {y(0)=y^{\prime}(0)=0} \\ {y=-3 e^{2 x}+4 e^{4 x}-12 x e^{2 x}-e^{-2 x}}\end{array}

OK. V.I.

4
##### Term Test 1 / Re: Problem 1 (morning)
« on: October 23, 2019, 07:42:27 AM »

\left(-y \sin (x)+y^{3} \cos (x)\right)+\left(3 \cos x+5 y^{2} \sin (x)\right) y^{\prime}=0

M _{y}=-\sin (x)+3 y^{2} \cos (x)

N_{ x}=-3 \sin (x)+5 y^{2} \cos (x)

M_{y} \neq N_{x}

so not exact

\begin{aligned} R_{1}=\frac{M_{ y}-N_{ x}}{M} &=\frac{-\sin (x)+3 y^{2} \cos (x)+3 \sin x-5 y^{2} \cos (x)}{-y \sin (x)+y^{3} \cos (x)} \\ &=\frac{2 \sin (x)-2 y^{2} \cos x}{-y \sin (x)+y^{3} \cos (x)}=-\frac{2}{y} \end{aligned}

u=e^{-2\int-\frac{1}{y} d y}=e^{2 \ln y}=y^{2}

so
\left(-y^{3} \sin (x)+y^{5} \cos (x)\right)+\left(3 y^{2} \cos (x)+5 y^{4} \sin (x)\right) y^{\prime}=0

\varphi_{x}=M

\varphi=\int M d x=y^{3} \cos (x)+y^{5} \sin (x)+h(x)

\varphi_{y}=N

\begin{aligned} \varphi _{y }&=3 y^{2} \cos (x)+5 y^{4} \sin (x)+h^{\prime}(x) \\ &=3 y^{2} \cos (x)+5 y^{4} \sin (x) \end{aligned}

so
h^{\prime}(x)=0

h(x)=c

so
\varphi_{\left(x, y\right)}=y^{3} \cos x+y^{5} \sin x=c

\item because
y\left(\frac{\pi}{4}\right)=\sqrt{2}

plug in

\begin{array}{c}{(\sqrt{2})^{3} \cos \frac{\pi}{4}+(\sqrt{2})^{5} \sin \frac{\pi}{4}=C} \\ {c=2+4=6}\end{array}

so

y^{3} \cos x+y^{5} \sin x=6

5
##### Term Test 1 / Re: Problem 2 (morning)
« on: October 23, 2019, 07:40:41 AM »
begin{equation}
\begin{array}{l}{y^{\prime \prime}-\frac{2 x+1}{x} y^{\prime}+(x+1) y=0} \\ {w=c e^{-f(x) d x}=C e^{b+\frac{1}{x} d x}=cx e^{2 x}}\end{array}

let $C=1$

w=\left|\begin{array}{cc}{e^{x}} & {y_{2}} \\ {e^{x}} & {y_{2}^{\prime}}\end{array}\right|=x e^{2 x}

\begin{array}{c}{e^{x} y^{\prime}-e^{x} y=x e^{2 x}} \\ {y^{\prime}-y=x e^{x}}\end{array}

u=e^{\int-1 d x}=e^{-x}

\begin{aligned} e^{-x} y^{\prime}-e^{-x} y &=x \\ e^{-x} y &=\int x d x \\ e^{-x} y &=\frac{x^{2}}{2}+c \end{aligned}

let $C=1$

y_{2}=\frac{x^{2} e^{x}}{2}+e^{x}

\item[c] so
\begin{array}{l}{y=c_{1} e^{x}+c_{2}\left(\frac{x^{2} e^{x}}{2}+e^{x}\right)} \\ {y^{\prime}=c_{1} e^{x}+c_{2}\left(\frac{2 x e^{x}}{2}+\frac{e^{x} x^{2}}{2}+e^{x}\right)}\end{array}

\begin{array}{l}{y(1)=0} \\ {y^{\prime}(1)=e}\end{array}

\begin{array}{l}{c_{1} e+G\left(\frac{e}{2}+e\right)=0} \\ {c_{1}+\frac{e}{2} c_{2}=0} \\ {c_{1} e+c_{2}\left(-e+\frac{e}{2}+e\right)= e} \\ {c_{1}+\frac{5}{2} c_{2}=1}\end{array}

\begin{array}{l}{C_{2}=1} \\ {C_{1}=-\frac{3}{2}}\end{array}

so

y=-\frac{3}{2} e^{x}+\left(\frac{x^{2}}{2} e^{-\frac{2}{2}}+e^{x}\right)

6
##### Quiz-4 / Quiz 4 TUT0202
« on: October 19, 2019, 04:46:15 PM »
4y'' + 12y' + 9y = 0
let y = e^rt
y' = re^rt
y''= r^2e^rt

4r^2 + 12r +9r = 0
r = -(2/3)
thus
y = c1e^(-(2/3))t + tc1e^(-(2/3))t

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