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### Messages - Aiting Zhang

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##### Quiz-5 / LEC0101 Quiz5
« on: November 01, 2019, 12:07:28 AM »
$$\mbox{Find a particular solution of the given non homogeneous equation.}$$
$$t^2{y}''+7t{y}'+5y=t, \quad t>0; \quad y_{1}(t)=t^{-1}$$
$$\mbox{Let} \ y=v(t)y_{1}(t)$$
$$\mbox{Then} \ {y}'={v}'t^{-1}-vt^{-2}$$
$${y}''={v}''t^{-1}-{v}'t^{-2}-{v}'t^{-2}+2vt^{-3}$$
$$\mbox{Plug y, y' and y'' in the given non homogeneous equation}$$
$$\mbox{Then} \ t^{2}({v}''t^{-1}-{v}'t^{-2}-{v}'t^{-2}+2vt^{-3})+7t({v}'t^{-1}-vt^{-2})+5vt^{-1}=t{v}''+5{v}'$$
$${v}''+5{v}'t^{-1}=1$$
$$\mbox{Let} \ r={v}', \ {r}'={v}''$$
$${r}'+5{r}'t^{-1}=1$$
$$p(t)=5t^{-1}$$
$$\mu=e^{\int\frac{5}{t} dt}=t^{5}$$
$$t^{5}{r}'+5t^{4}r=t^{5}$$
$$r=\frac{1}{6}t+c_{1}t^{-5}={v}'$$
$$\mbox{Then} \ v=\int\frac{1}{6}t+c_{1}t^{-5}dt$$
$$v=\frac{1}{12}t^2-\frac{1}{4}c_{1}t^{-4}+c_{2}$$
$$\mbox{Thus,} \ y=\frac{1}{12}t^{2}t^{-1}-\frac{1}{4}c_{1}t^{-4}t^{-1}+c_{2}t^{-1}$$
$$y=\frac{1}{12}t-\frac{c_{1}}{4}t^{-5}+c_{2}t^{-1}$$
$$\mbox{Since} \ \frac{-c_{1}}{4} \mbox{is a constant. Therefore, the particular solution is } \ y=\frac{1}{12}t+c_{1}t^{-5}+c_{2}t^{-1}$$

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##### Quiz-4 / TUT0302 Quiz4
« on: October 18, 2019, 02:00:05 PM »
$$\mbox{Use the Euler equation method to solve the given equation for t > 0}$$
$$t^2{y}''-4t{y}'+6y=0$$
$$\frac{d^2y}{dx^2}+(\alpha-1)\frac{dy}{dx}+\beta y=0$$
$$\frac{d^2y}{dx^2}+(-4-1)\frac{dy}{dx}+6y=0$$
$$\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=0$$
$$r^2-5r+6=0$$
$$(r-2)(r-3)=0$$
$$r_{1}=2 \quad r_{2}=3$$
$$\mbox{Thus, } y(x)=c_{1}e^2x+c_{2}e^3x$$
$$\mbox{Substituting }x=ln(t)$$
$$y(t)=c_{1}e^2ln(t)+c_{2}e^3ln(t)$$
$$y(t)=c_{1}e^{ln(t)^2}+c_{2}e^{ln(t)^3}$$
$$\mbox{Therefore, the solution is } y(t)=c_{1}t^2+c_{2}t^3$$

3
##### Quiz-3 / TUT0302 Quiz3
« on: October 11, 2019, 02:00:00 PM »
$$\mbox{Find the solution of the given initial value problem }$$
$${2y}''+{y}'-4y=0$$
$$y\left(0 \right)=1 \mbox{ and } {y}'\left(0 \right)=0$$
$$2r^2+r-4=0$$
$$r=\frac{-1\pm\sqrt{1-4\times2\times(-4)}}{4}=\frac{-1\pm\sqrt{33}}{4}$$
$$\mbox{Hence, }r_{1}=\frac{-1+\sqrt{33}}{4} \mbox{ and } r_{2}=\frac{-1-\sqrt{33}}{4}$$
$$\mbox{Then the general solution of the given differential equation is }$$
$$y=c_{1}e^{\frac{-1+\sqrt{33}}{4} t}+c_{2}e^{\frac{-1-\sqrt{33}}{4} t}$$
$$\because y\left(0 \right)=1$$
$$\therefore c_{1}+c_{2}=1$$
$${y}'=\frac{-1+\sqrt{33}}{4}c_{1}e^{\frac{-1+\sqrt{33}}{4} t}+\frac{-1-\sqrt{33}}{4}c_{2}e^{\frac{-1-\sqrt{33}}{4} t}$$
$$\because {y}'\left(0 \right)=0$$
$$\therefore \frac{-1+\sqrt{33}}{4}c_{1}+\frac{-1-\sqrt{33}}{4}c_{2}=0$$
$$\mbox{Since }c_{1}+c_{2}=1, \mbox{so }c_{1}=1-c_{2}$$
$$\therefore \frac{-1+\sqrt{33}}{4}(1-c_{2})+\frac{-1-\sqrt{33}}{4}c_{2}=0$$
$$\frac{-1+\sqrt{33}}{4}-c_{2}\frac{2\sqrt{33}}{4}=0$$
$$\therefore c_{2}=\frac{33-\sqrt{33}}{66}$$
$$c_{1}=1-\frac{33-\sqrt{33}}{66}=\frac{33+\sqrt{33}}{66}$$
$$\mbox{Therefore, the solution of the given initial value problem is }$$
$$y=\frac{33+\sqrt{33}}{66}e^{\frac{-1+\sqrt{33}}{4} t}+\frac{33-\sqrt{33}}{66}e^{\frac{-1-\sqrt{33}}{4} t}$$

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##### Quiz-2 / TUT0302 Quiz2
« on: October 04, 2019, 02:00:01 PM »
$$\mbox{Solve the equation }$$
$$x^2y^3+x(1+y^2)y'=0\qquad\mu (x,y)=\dfrac{1}{xy^3}$$
$$\mbox{Define }M(x,y)=x^2y^3,\quad N(x,y)=x(1+y^2)$$
$$M_y=\dfrac{\partial}{\partial y}[x^2y^3]=3x^2y^2$$
$$N_x=\dfrac{\partial}{\partial x}[x(1+y^2)]=1+y^2$$
$$\mbox{Since }M_y\neq N_x, \mbox{this implies the given equation is not exact. Let's show the given equation multiplied by the integrating factor } \mu (x,y)=\dfrac{1}{xy^3} \mbox{is exact}.$$
$$\dfrac{1}{xy^3}x^2y^3+\dfrac{1}{xy^3}x(1+y^2)y'=x+(y^{-3}+y^{-1})y'=0$$
$$\mbox{Define }M'(x,y)=x,N'(x,y)=y^{-3}+y^{-1}$$
$$M'_y=\dfrac{\partial}{\partial y}{(x)}=0$$
$$M'_x=\dfrac{\partial}{\partial x}[y^{-3}+y^{-1}]=0$$
$$\mbox{Since }M'_y=M'_x, \mbox{this implies } x+(y^{-3}+y^{-1})y'=0 \mbox{ is exact}. \mbox{Thus, there exists a function } \varphi (x,y)$$
$$\mbox{such that}\quad \varphi_x=M, \varphi_y=N$$
\begin{equation*}
\begin{aligned}
\varphi &=\int M\ dx \\
&=\int x\ dx \\
&=\dfrac{1}{2}x^2+h(y)
\end{aligned}
\end{equation*}
$$\varphi_y=h'(y)=y^{-3}+y^{-1}$$
$$h(y)=-\dfrac{1}{2}y^{-2}+ln|y|+C$$
$$\mbox{Therefore},\dfrac{1}{2}x^2-\dfrac{1}{2}y^{-2}+ln|y|=C \mbox{ is the general solution for the given equation}.$$

5
##### Quiz-1 / TUT0302 Quiz1
« on: September 27, 2019, 02:00:17 PM »
$\frac{d y}{d x}=\frac{x^{2}-3 y^{2}}{2 x y}$
$u=\frac{y}{x} \quad y=u \cdot x$
$\frac{d y}{d x}=\frac{d(u \cdot x)}{d x}=\frac{d u}{d x} \cdot x+u \cdot 1$
$\frac{d y}{d x}=\frac{1-\frac{3 y^{2}}{x^{2}}}{\frac{2 x y}{x^2}}=\frac{1-\frac{3 y^{2}}{x^{2}}}{\frac{2 y}{x}}$
$\frac{d y}{d x}=\frac{1-3 u^{2}}{2 u}$
$\frac{d u}{d x} \cdot x+u=\frac{1-3 u^{2}}{2 u}$
$\frac{d u}{d x} \cdot x=\frac{1-5 u^{2}}{2 u}$
$\int \frac{2 u}{1-5 u^{2}} d u=\int \frac{1}{x} d x$
$-\frac{1}{5}\ln| 1-5 u^{2}|=\ln | x |+c$
$e^{-\frac{1}{5} \ln \left|1-5 u^{2}\right|}=e^{\ln |x|+c}$
$e^{\ln \left|1-5 u^{2}\right|^{-\frac{1}{5}}}=e^{\ln |x |} \cdot e^{c}$
$x e^{c}=\left(1-5 u^{2}\right)^{-\frac{1}{5}}$
$x e^{c}=\left(1-5 \frac{y^{2}}{x^{2}}\right)^{-\frac{1}{5}}$

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