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### Messages - Anyue Huang

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1
##### Term Test 2 / Re: Problem 1 (main sitting)
« on: November 19, 2019, 05:24:49 AM »
Solution see attachment.

2
##### Quiz-5 / TUT0202 Quiz5
« on: November 01, 2019, 02:00:08 PM »
Verify that the given functions y 1and y2 satisfy the corresponding homogeneous equation, then find a particular solution of the given nonhomogeneous equation.
$$t^{2} y^{\prime \prime}-t(t+2) y^{\prime}+(t+2) y=2 t^{3}, t>0 ; y_{1}(t)=t, y_{2}(t)=t e^{t}$$

$$y_{1}(t)=t \quad y_{1}^{\prime}(t)=1 \quad y_{1}^{\prime \prime}(t)=0$$
$$-t(t+2)+(t+2) t=0$$
$$y_{2}(t)=t e^{t} \quad y_{2}^{\prime}(t)=e^{t}+t e^{t} \quad y_{2}^{\prime \prime}(t)=2 e^{t}+t e^{t}$$
$$t^{2}\left(2 e^{t}+t e^{t}\right)-t(t+2)\left(e^{t}+t e^{t}\right)+(t+2) t e^{t}=0$$

Hence, $y_{1}$ and $y_{2}$ satisfy the homogeneons equation
$$y^{\prime \prime}-\frac{t+2}{t} y^{\prime}+\frac{t+2}{t^{2}} y=2 t$$
\begin{aligned} w&=\left|\begin{array}{cc}{t} & {t e^{t}} \\ {1} & {e^{t}+\operatorname{te}^{t}}\end{array}\right|=t^{2} e^{t}\\ w_{1}&=\left|\begin{array}{cc}{0} & {t e^{t}} \\ {1} & {e^{t}+t e^{t}}\end{array}\right|=-t e^{t}\\ w_{2}&=\left|\begin{array}{cc}{t} & {0} \\ {1} & {1}\end{array}\right|=t \end{aligned}
\begin{aligned} y_{p}(t) &=t \int \frac{\left(-t e^{t}\right)(2 t)}{t^{2} e^{t}} d t+t e^{t} \int \frac{(t)(2 t)}{t^{2} e^{t}} d t \\ &=t \int-2 d t+2 t e^{t} \int \frac{1}{e^{t}} d t \\ &=t(-2 t)+2 t e^{t}\left(-e^{-t}\right) \\ &=-2 t^{2}-2 t \end{aligned}

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##### Quiz-5 / TUT0202 Quiz5
« on: November 01, 2019, 09:57:46 AM »
Verify that the given functions y 1and y2 satisfy the corresponding homogeneous equation;
then find a particular solution of the given nonhomogeneous equation.
$$t^{2} y^{\prime \prime}-t(t+2) y^{\prime}+(t+2) y=2 t^{3}, t>0 ; y_{1}(t)=t, y_{2}(t)=t e^{t}$$

$$y_{1}(t)=t \quad y_{1}^{\prime}(t)=1 \quad y_{1}^{\prime \prime}(t)=0$$
$$-t(t+2)+(t+2) t=0$$
$$y_{2}(t)=t e^{t} \quad y_{2}^{\prime}(t)=e^{t}+t e^{t} \quad y_{2}^{\prime \prime}(t)=2 e^{t}+t e^{t}$$
$$t^{2}\left(2 e^{t}+t e^{t}\right)-t(t+2)\left(e^{t}+t e^{t}\right)+(t+2) t e^{t}=0$$

Hence, $y_{1}$ and $y_{2}$ satisfy the homogeneons equation
$$y^{\prime \prime}-\frac{t+2}{t} y^{\prime}+\frac{t+2}{t^{2}} y=2 t$$
\begin{aligned} w&=\left|\begin{array}{cc}{t} & {t e^{t}} \\ {1} & {e^{t}+\operatorname{te}^{t}}\end{array}\right|=t^{2} e^{t}\\ w_{1}&=\left|\begin{array}{cc}{0} & {t e^{t}} \\ {1} & {e^{t}+t e^{t}}\end{array}\right|=-t e^{t}\\ w_{2}&=\left|\begin{array}{cc}{t} & {0} \\ {1} & {1}\end{array}\right|=t \end{aligned}
\begin{aligned} y_{p}(t) &=t \int \frac{\left(-t e^{t}\right)(2 t)}{t^{2} e^{t}} d t+t e^{t} \int \frac{(t)(2 t)}{t^{2} e^{t}} d t \\ &=t \int-2 d t+2 t e^{t} \int \frac{1}{e^{t}} d t \\ &=t(-2 t)+2 t e^{t}\left(-e^{-t}\right) \\ &=-2 t^{2}-2 t \end{aligned}

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##### Term Test 1 / Re: Problem 3 (main)
« on: October 23, 2019, 08:04:31 AM »
Find the general solution of the equation
$$y^{\prime \prime}-2 y^{\prime}-3 y=16 \cosh x$$
\begin{aligned} r^{2}-2 r-3 &=0 \\(r-3)(r+1) &=0 \\ r=3, r&=-1 \end{aligned}
$$\begin{array}{l}{y_{c}=c_{1} e^{3 x}+c_{2} e^{-x}} \\ {y^{\prime \prime}-2 y^{\prime}-3 y=16 \cosh x=8 e^{x}+8 e^{-x}} \\ {y^{\prime \prime}-2 y^{\prime}-3 y=8 e^{x}}\end{array}$$

Let $y_{p}=A e^{x}\quad y^{\prime}=A e^{x} \quad y^{\prime \prime}=A e^{x}$
\begin{aligned} A e^{x}-2 A e^{x}-3 A e^{x} &=8 e^{x} \\-4 A e^{x} &=8 e^{x} \\ A &=-2 \end{aligned}
$$\therefore y_{p}=-2 e^{x}$$

$$y^{\prime \prime}-2 y^{\prime}-3 y=8 e^{-x}$$

$\operatorname{let} y_{p}(t)=A x e^{-x} \quad y^{\prime}=A e^{-x}-A x e^{-x}$
\begin{aligned} y^{\prime \prime}=&-A e^{-x}-\left(A e^{-x}-A x e^{-x}\right) \\ &=-2 A e^{-x}+A x e^{-x} \end{aligned}
$$-2 A e^{-x}+A x e^{-x}-2 A e^{-x}+2 A x e^{-x}-3 A x e^{-x}=8 e^{-x}$$
\begin{aligned}-2 A-2 A &=8 \\ A &=-2 \\ y_{p} &=-2 \times e^{-x} \end{aligned}
$$y=c_{1} e^{3 x}+c_{2} e^{-x}-2 e^{x}-2 x e^{-x}$$

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##### Quiz-3 / TUT0202 Quiz3
« on: October 11, 2019, 02:00:30 PM »
Find a differential equation whose general solution is $y=c_{1} e^{-t / 2}+c_{2} e^{-2 t}$.

Then $r_{1}=-\frac{1}{2}, r_{2}=-2$ are two roots of the characteristic equation of the required differential equation.

so the characteristic equation should look like
\begin{aligned}\left(r+\frac{1}{2}\right)(r+2) &=0 \\(2 r+1)(r+2) &=0 \\ 2 r^{2}+5 r+2 &=0 \end{aligned}

which corresponds to the DE

$$2 y^{\prime \prime}+5 y^{\prime}+2 y=0$$

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