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### Messages - Yuanxi Gong

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##### Quiz-5 / Quiz5 Victor's section Lec5101
« on: November 01, 2019, 02:00:01 PM »
$$y^{\prime \prime}+4 y^{\prime}+4 y=t^{-2} e^{-2 t} \quad t>0$$

$\begin{array}{c}{r^{2}+4 r+4=0} \\ {(r+2)^{2}=0} \\ {r=-2,-2}\end{array}$

Since the roots of characteristic equation are real and repeating.

Therefore, the complimentary solution of differential equation is as follows:

$y_{e}(t)=c e^{-2 t}+c_{2} t e^{-2 t}$

From complimentary solution, the two fundamental solutions of the differential equation are
$y_{1}(t)=e^{-2 t}$ and $y_{2}(t)=t e^{-2 t}$.

Apply the method of variation of parameters to determine the particular solution $Y(t)$
Determine the Wronskian as follows:

Determine the Wronskian as follows:

\begin{aligned} W\left(y_{1}, y_{2}\right)(t) &=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{-2 t}} & {y_{2}^{\prime}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{-2 t}} & {y_{2}^{\prime}(t)}\end{array}\right| \\ &=e^{-2 t}\left(-2 t e^{-2 t}+e^{-2 t}\right)-\left(-2 e^{-2 t}\right)\left(t e^{-2 t}\right) \\ &=-2 t e^{-t^{4}}+e^{-4 t}+t e^{-4 t} \end{aligned}

$W\left(y_{1}, y_{2}\right)(t)=e^{-4 t}$

By the method of variation of parameters, the particular solution is given as follows:

$Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)$

Here, $u_{1}(t)$ and $u_{2}(t)$ are the parameters defined as follows:

$u_{1}(t)=-\int \frac{y_{2}\left(t^{2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t$ and $u_{2}(t)=\int \frac{y_{1}\left(t^{2} e^{-t^{2}}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t$

\begin{aligned} u_{1}(t) &=-\int \frac{y_{2}\left(r^{-2} e^{-z}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t \\ &=-\int \frac{t e^{-2 t}-f^{2} e^{-z t}}{e^{-4}} d t \\ &=-\int \frac{r^{-2} e^{-t}}{e^{-t}} d t \\ &=-\int t^{-1} d t \\=&-\ln t \end{aligned} \begin{aligned} u_{2}(t) &=\int \frac{y_{1}\left(t^{-2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t \\ &=\int \frac{e^{-2 t} t^{-2} e^{-t}}{e^{-4 t}} d t \\ &=\int \frac{e^{-t^{2} t^{-2}}}{e^{-4 t}} d t \\ &=\int t^{-2} d t \\ &=-t^{-1} \end{aligned}

Substitute all values in equation $Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)$ for particular solution.

\begin{aligned} Y(t) &=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t) \\ &=(-\ln t)\left(e^{-2 t}\right)+\left(t e^{-2 t}\right)\left(-t^{-1}\right) \\ &=-e^{-2 t} \ln t-e^{-2 t} \end{aligned}

Add the complimentary solution and particular solution to determine the general solution of the
differential equation as follows:

\begin{aligned} y &=y_{e}(t)+Y(t) \\ &=c e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t-e^{-2 t} \\ &=(c-1) e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t \\ &=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t \end{aligned}

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##### Quiz-4 / TUt0701 quiz4
« on: October 18, 2019, 02:34:12 PM »
𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛:𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑦″+4𝑦′+5𝑦=0   y(0)=1 , 𝑦′(0)=0
r^2 + 4r +5 =0
r = (-4±√16-20)/2
r = -2±i
y = C1e^(-2t)sint + C2e^(-2t)cost
initial condition: y(0)=1 , 𝑦′(0)=0
C2 * 1 = 1
C2 = 1
𝑦′ = -2C1e^(-2t)sint + C1e^(-2t)cost + -2C2e^(-2t)cost - C1e^(-2t)sint
C1 -2C2 = 0
C1 = 2
y = 2e^(-2t)sint + e^(-2t)cost

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##### Quiz-2 / TUT0701 quiz2
« on: October 04, 2019, 02:26:16 PM »
Find an integrating factor and solve the given equation  (3x + 6/y) + (x^2/y + 3y/x)dy/dx =0
(3x + 6/y)dx + (x^2/y + 3y/x)dy =0
𝑀𝑦 = -6/y^2  N𝑥=2x/y -3y/x^2  not exact
(N𝑥-𝑀𝑦) / (x*M-N*y) = (2x/y - 3y/x^2 + 6/y^2)/(2x^2 + 6x/y - 3y^2/x) = 1/xy
Let t represent xy
μ = e^∫1/t𝑑t = e^ln|t| = t = xy
multiply xy on both side
(3yx^2 + 6x)dx + (x^3 +3y^2)dy = 0
𝜙(𝑥,𝑦)=∫3yx^2 + 6x𝑑𝑥=yx^3 + 3x^2 + ℎ(𝑦)
𝜙(𝑦)=x^3 + ℎ′(𝑦) = N
ℎ′(𝑦)=3y^2
ℎ(𝑦) = y^3
𝜙(𝑥,𝑦)=yx^3 + 3x^2 + y^3 = c

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