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### Topics - Zuwei Zhao

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##### Quiz-5 / TUT 5103 Quiz5
« on: November 01, 2019, 02:00:00 PM »
\noindent Verify that the given functions y 1and y 2 satisfy the corresponding homogeneous equation;
then find a particular solution of the given nonhomogeneous equation.
$$\begin{array}{l}{(1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, 0<t<1} \\ {y_{1}(t)=e^{t}, y_{2}(t)=t}\end{array}$$

$$y^{\prime \prime}+\frac{t}{1-t} y^{\prime}-\frac{1}{1-t} y=-2(t-1) e^{-t}$$
$$\begin{array}{l}{w=\left|\begin{array}{ll}{e^{t}} & {t} \\ {e^{t}} & {1}\end{array}\right|=e^{t}-t e^{t}} \\ {w_{1}=\left|\begin{array}{ll}{0} & {t} \\ {1} & {1}\end{array}\right|=-t \quad w_{2}=\left|\begin{array}{ll}{e^{t}} & {0} \\ {e^{t}} & {1}\end{array}\right|=e^{\tau}}\end{array}$$
\begin{aligned} Y(t) &=e^{t} \int \frac{-t \cdot\left(-2(t-1) e^{-t}\right)}{e^{t}-t e^{t}} d t+t \int \frac{e^{t}\left(-2(t-1) e^{-t}\right)}{e^{t}-t e^{t}} d t \\ &=e^{t} \int \frac{-t \cdot\left(t^{2}(t-1) e^{-t}\right)}{e^{t}(1-t)}+t \int \frac{e^{t}\left(t+2(t-1) e^{-t}\right)}{e^{t}(1-t)} d t \\ &=e^{t} \int-2 t e^{-2 t} d t+\int 2 e^{-t} d t \\ &=-2 e^{t} \int t e^{-2 t} d t-2 t e^{-t} \end{aligned}
$$\begin{array}{ll}{u=t} & {d v=e^{-2 t}} \\ {d u=d t} & {v=-\frac{1}{2} e^{-2 t}}\end{array}$$
$$\begin{array}{l}{\displaystyle =-2 e^{t}\left(-\frac{1}{2} e^{-2 t} \cdot t+\int \frac{1}{2} e^{-2 t} d t\right)+t \int 2 e^{-t} d t} \\ {\displaystyle =-2 e^{t}\left(-\frac{1}{2} e^{-2 t} t-\frac{1}{4} e^{-2 t}\right)-2 t e^{-t}} \\ {\displaystyle =t e^{-t}+\frac{1}{2} e^{-t}-2 t e^{-t}} \\ {\displaystyle =\left(\frac{1}{2}-t\right) e^{-t}}\end{array}$$

$\therefore$ the particular solution of the given non-homogeneous equation is
$$Y(t)=\left(\frac{1}{2}-t\right) e^{-t}$$

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##### Quiz-4 / TUT 5103 Quiz4
« on: October 18, 2019, 02:00:02 PM »
$$\begin{array}{l}{y^{\prime \prime}+y=0, \quad y\left(\frac{\pi}{3}\right)=2, \quad y^{\prime}\left(\frac{\pi}{3}\right)=-4} \\ {r^{2}+1=0}\end{array}$$
\begin{aligned} r &=\frac{-0 \pm \sqrt{0-4 \cdot 1} \cdot 1}{2} \\ &=\frac{-0+2 i}{2} \\ &=0 \pm i \end{aligned}
$$\begin{array}{l}{y=c_{1} e^{o t} \cos t+c_{2} e^{o t} \sin t} \\ {y=c_{1} \cos t+c_{2} \sin t} \\ {y^{\prime}=-c_{1} \sin t+c_{2} \cos t}\end{array}$$
$$\left\{\begin{array}{l}{2=\frac{1}{2} c_{1}+\frac{\sqrt{3}}{2} c_{2}} \\ {-4=-\frac{\sqrt{3}}{2} c_{1}+\frac{1}{2} c_{2}}\end{array}\right.$$
\begin{aligned} c_{1} &=1+2 \sqrt{3}\quad c_{2}=\sqrt{3}-2 \\ \therefore y &=(1+2 \sqrt{3}) \cos t+(\sqrt{3}-2) \sin t \end{aligned}

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##### Quiz-3 / MAT244TUT5103 Quiz3
« on: October 11, 2019, 02:00:03 PM »
Find the Wronskian of the given pair of functions.
$$x, x e^{x}$$

$$\begin{array}{l}{(x)^{\prime}=1} \\ {\left(x e^{x}\right)^{\prime}=x e^{x}+e^{x}}\end{array}$$
$$W=\left|\begin{array}{cc}{x} & {x e^{x}} \\ {1} & {x e^{x}+e^{x}}\end{array}\right|=x^{2} e^{x}+x e^{x}-x e^{x}=x^{2} e^{x}$$

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##### Quiz-2 / MAT24f4 TUT5103 Quiz2
« on: October 04, 2019, 02:02:30 PM »
$$(ye^{2xy}+x)+bxe^{2xy}y^{\prime}=0.$$
\begin{aligned} M&=ye^{2xy}+x\\ M_y&=e^{2xy}+2xye^{2xy}\\ N_x&=bxe^{2xy}+2bxye^{2xy}\\ e^{2xy}&+2xye^{2xy}=bxe^{2xy}+2bxye^{2xy}\\ &b=1\\ \end{aligned}

Substitute $(ye^{2xy}+x)+xe^{2xy}y^{\prime}=0$

\begin{aligned} M&=ye^{2xy}+x\qquad N=xe^{2xy}\\ M_y&=N_x\qquad \therefore \text{the function now is exact.}\\ \varphi_x&=M\\ \varphi&=\int ye^{2xy}+xdx\\ &=\frac{ye^{2xy}}{2y}+\frac{x^2}{2}+h(y)=\frac{e^{2xy}}{2}\frac{x^2}{2}+h(y)\\ \varphi_y&=\frac{2xe^{2xy}}{2}+h^{\prime}(y)\\ &=xe^{2xy}+h^{\prime}(y)\\ h^{\prime}(y)&=0\\ h(y)&=\text{constant}\\ \varphi&=\frac{e^{2xy}}{2}+\frac{x^2}{2}=c \end{aligned}

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##### Quiz-2 / MAT24f4 TUT5103 Quiz2
« on: October 04, 2019, 02:01:16 PM »
$$(ye^{2xy}+x)+bxe^{2xy}y^{\prime}=0.$$
\begin{aligned} M&=ye^{2xy}+x\\ M_y&=e^{2xy}+2xye^{2xy}\\ N_x&=bxe^{2xy}+2bxye^{2xy}\\ e^{2xy}&+2xye^{2xy}=bxe^{2xy}+2bxye^{2xy}\\ &b=1\\ \end{aligned}

Substitute $(ye^{2xy}+x)+xe^{2xy}y^{\prime}=0$

\begin{aligned} M&=ye^{2xy}+x\qquad N=xe^{2xy}\\ M_y&=N_x\qquad \therefore \text{the function now is exact.}\\ \varphi_x&=M\\ \varphi&=\int ye^{2xy}+xdx\\ &=\frac{ye^{2xy}}{2y}+\frac{x^2}{2}+h(y)=\frac{e^{2xy}}{2}\frac{x^2}{2}+h(y)\\ \varphi_y&=\frac{2xe^{2xy}}{2}+h^{\prime}(y)\\ &=xe^{2xy}+h^{\prime}(y)\\ h^{\prime}(y)&=0\\ h(y)&=\text{constant}\\ \varphi&=\frac{e^{2xy}}{2}+\frac{x^2}{2}=c \end{aligned}

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##### Quiz-2 / MAT24f4 TUT5103 Quiz2
« on: October 04, 2019, 02:00:45 PM »
$$(xy^{2}+bx^{2}y)+(x+y)x^2y^{\prime}=0$$
\begin{aligned} M&=xy^2+bx^2y\qquad N=(x+y)x^{2}=x^3+yx^2\\ M_y&=2yx+bx^2\qquad N_x=3x^2+2yx\\ \therefore &b=3 \end{aligned}
$$(xy^{2}+3x^{2}y)+(x+y)x^2y^{\prime}=0$$
\begin{aligned} M&=xy^2+3x^2y\qquad N=(x+y)x^{2}=x^3+yx^2\\ M_y&=2yx+3x^2\qquad N_x=3x^2+2yx\\ \therefore &M_y=N_x\quad\therefore \text{the function is exact.} \end{aligned}
\begin{aligned} \exists ~\text{s.t.} \varphi_x&=M\\ \varphi&=\int M d x\\ &=\int xy^2+3x^2ydx\\ &=\frac{x^2y^2}{2}+x^3y+h(y)\\ \varphi_y&=\frac{2yx^2}{2}+x^3+h^{\prime}(y)\\ &=x^2y+x^3=h^{\prime}(y)\\ &=x^3+x^2y\\ h^{\prime}(y)&=0\\ h(y)&=\text{constant}\\ \varphi&=\frac{x^2y^2}{2}+x^3y=c \end{aligned}

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##### Quiz-2 / MAT24f4 TUT5103 Quiz2
« on: October 04, 2019, 02:00:11 PM »
$$(ye^{2xy}+x)+bxe^{2xy}y^{\prime}=0.$$
\begin{aligned} M&=ye^{2xy}+x\\ M_y&=e^{2xy}+2xye^{2xy}\\ N_x&=bxe^{2xy}+2bxye^{2xy}\\ e^{2xy}&+2xye^{2xy}=bxe^{2xy}+2bxye^{2xy}\\ &b=1\\ \end{aligned}

Substitute $(ye^{2xy}+x)+xe^{2xy}y^{\prime}=0$

\begin{aligned} M&=ye^{2xy}+x\qquad N=xe^{2xy}\\ M_y&=N_x\qquad \therefore \text{the function now is exact.}\\ \varphi_x&=M\\ \varphi&=\int ye^{2xy}+xdx\\ &=\frac{ye^{2xy}}{2y}+\frac{x^2}{2}+h(y)=\frac{e^{2xy}}{2}\frac{x^2}{2}+h(y)\\ \varphi_y&=\frac{2xe^{2xy}}{2}+h^{\prime}(y)\\ &=xe^{2xy}+h^{\prime}(y)\\ h^{\prime}(y)&=0\\ h(y)&=\text{constant}\\ \varphi&=\frac{e^{2xy}}{2}+\frac{x^2}{2}=c \end{aligned}

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