Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.


Topics - Joy Zhou

Pages: [1]
1
Quiz-5 / LEC5101 Quiz5
« on: November 01, 2019, 02:12:58 PM »
$y''+4y'+4y=t^{-2}e^{-2t}, t>0$

The homogeneous differential equation is as follows:
$$y^{\prime\prime}+4 y^{\prime}+4 y=0$$

The characteristic equation for homogeneous differential equation is
$$\begin{aligned}
r^{2}+4 r+4 &=0\\
(r+2)^{2} &=0 \\ r &=-2,-2 \end{aligned}
$$

since the roots of characteristic equation are real and repeating.

Therefore, the solution of differential equation is as follows:
$$
{y_{c}(t)=c e^{-2 t}+c_{2} te^{-2t}}
$$

($c$ and $c_{2}$ are constants)

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

Then, use the variation of parameters method to determine the particular solution $Y(t) .$

Find the Wronskian as follows:
$$\begin{aligned} W\left(y_{1}, y_{2}\right)(t) &=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{-2 t}} & {te^{-2t}} \\ {-2 e^{-2 t}} & {-2 t e^{-2 t}+e^{-2 t}}\end{array}\right| \\ &=e^{-2 t}\left(-2 t e^{-2 t}+e^{-2 t}\right)-\left(-2 e^{-2 t}\right)\left(t e^{-2 t}\right) \\ &=-2 t e^{-4 t}+e^{-4 t}+t e^{-4 t} \end{aligned}$$

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

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

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

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

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

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

So, the general solution of the equation is:
$$
\begin{array}{l}{y=y_{c}(t)+Y(t)} \\ {=c e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t-e^{-2 t}} \\ {=(c-1) e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t} \\ {=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t}\end{array}
$$

$$(c-1=c_1)
$$
Therefore, the required general solution is
$$
y(t)=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t
$$

2
Quiz-4 / TUT0601 Quiz4
« on: October 18, 2019, 09:44:50 PM »
Find the general solution of given equation:
$$y^{\prime \prime}+y^{\prime}-6 y=12 e^{3 t}+12 e^{-2 t}$$

Step1. To find the general solution, get the homogeneous equation
$$y^{\prime \prime}+y^{\prime}-6 y=0$$

The characteristic equation is
$$r^{2}+r-6=0$$
$$(r+3)(r-2)=0$$
$$r=2, -3$$

Thus, the solution of homogeneous differential equation $y^{\prime \prime}+y^{\prime}-6 y=0$ is
$$y_{c}(t)=c_{1} e^{-3 t}+c_{2} e^{2 t}$$

Step2. To find the particular solution, use undetermined coefficients method.
Suppose $y_{1}(t)=A e^{3_{t}}$ is a function satisfying the equation
$$y_{1}^{\prime \prime}+y_{1}^{\prime}-6 y_{1}=12 e^{3t}$$

Then,
$$
\begin{array}{l}{y_{1}(t)=A e^{3 t}} \\ {y_{1}^{\prime}(t)=3 A e^{3 t}} \\ {y_{1}^{\prime \prime}(t)=9 A e^{3 t}}\end{array}
$$
$$9 A e^{3 t}+3 A e^{3 t}-6 A e^{3 t}=12 e^{3 t}$$
$$6 A e^{3 t}=12 e^{3 t}$$
$$A=2$$
So, the solution for $y_{1}(t)$ is
$$y_{1}(t)=2 e^{3 t}$$

Now suppose $y_{2}(t)=B e^{-2 t}$ to satisfies the equation
$$y_{1}^{\prime \prime}+y_{1}^{\prime}-6 y_{1}=12 e^{-2 t}$$

Then,
$$y_{2}(t)=B e^{-2 t}$$
$$y_{2}^{\prime}=-2 B e^{-2 t}$$
$$y_{2}^{\prime \prime}=4 B e^{-2 t}$$
$$4 B e^{-2 t}-2 B e^{-2 t}-6 B e^{-2 t}=12 e^{-2 t}$$
$$-4 B e^{-2 t}=12 e^{-2 t}$$
$$B=-3$$

So, the solution for $y_{2}(t)$ is
$$y_{2}(t)=-3 e^{-2 t}$$

Therefore, the general solution for the non-homogeneous differential equation is
$$y=y_{c}(t)+y_{1}(t)+y_{2}(t)$$
$$
y=c_{1} e^{-3 t}+c_{2} e^{2 t}+2 e^{3 t}-3 e^{-2 t}
$$

3
Quiz-3 / TUT0601 Quiz3
« on: October 11, 2019, 09:47:00 PM »
Find the Wronskian of two solutions of the given differential equation without solving the
equation.
$$
\cos (t) y^{\prime \prime}+\sin (t) y^{\prime}-t y=0
$$

First, we divide both sides of the equation by $\cos (t):$
$$
y^{\prime \prime}+\tan (t) y^{\prime}-\frac{t}{\cos (t)} y=0
$$

Now the given second-order differential equation has the form:
$$
L[y]=y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0
$$

Noting if we let $p(t)=\tan (t)$ and $q(t)=-\frac{t}{\cos (t)},$ then $p(t)$ is continuous
everywhere except at $\frac{\pi}{2}+k \pi,$ where $k=0,1,2, \ldots$ and $q(t)$ is also continuous
everywhere except at $t=0$.

Therefore, by Abel's Theorem: the Wronskian $W\left[y_{1}, y_{2}\right](t)$ is given by
$$
\begin{aligned} W\left[y_{1}, y_{2}\right](t) &=cexp\left(-\int p(t) d t\right) \\ &=cexp\left(-\int \tan (t) d t\right) \\ &=c e^{\ln |\cos (t)|} \\ &=c\cos (t) \end{aligned}
$$

4
Quiz-2 / TUT0601 Quiz2
« on: October 05, 2019, 06:31:35 PM »
Find an integrating factor and solve the given equation.
$$
\left(3 x+\frac{6}{y}\right)+\left(\frac{x^{2}}{y}+3 \frac{y}{x}\right) \frac{d y}{d x}=0
$$

We want to find an integrating factor $\mu$ as a function of $xy$ such that
$(\mu M)_{y}=(\mu N)_{x}$, Let $z=xy$. Thus, $\mu(x y)=\mu(z(x, y))$ Then

$
\mu_{x}(x y)=\frac{d \mu}{d z} \frac{\partial z}{\partial x}=y \frac{d \mu}{d z}
$
$\mu_{y}(x y)=\frac{d \mu}{d z} \frac{\partial z}{\partial y}=x \frac{d \mu}{d z}$

Therefore,
$$
(\mu M)_{y}=(\mu N)_{x}
$$
$$
\mu M_{y}+x M \frac{d \mu}{d z}=\mu N_{x}+y N \frac{d \mu}{d z}
$$
$$
\mu\left(M_{y}-N_{x}\right)=\frac{d \mu}{d z}(y N-x M)
$$
$$
\frac{\mathrm{d} \mu}{\mathrm{d} z}=\mu\left(\frac{N_{x}-M_{y}}{x M-y N}\right)
$$

Therefore,

$
\mu(z)=\exp \left(\int R(z) \mathrm{d} z\right)
$
\quad where $R(z)=R(x y)=\frac{N_{x}-M_{y}}{x M-y N}$

Returning to our original differential equation, let

$M(x, y)=3 x+\frac{6}{y}$ \quad and \quad $N(x, y)=\frac{x^{2}}{y}+3 \frac{y}{x}=0$

Then

$\frac{\partial}{\partial y} M(x, y)=\frac{-6}{y^{2}}$ \quad and \quad $\frac{\partial}{\partial x} N(x, y)=\frac{2 x}{y}-\frac{3 y}{x^{2}}$

We can see that this equation is not exact, however, note that
$$\frac{N_{x}-M_{y}}{x M-y N}=\frac{\frac{2 x}{y}-\frac{3 y}{x^{2}}+\frac{6}{y^{2}}}{x\left(3 x+\frac{6}{y}\right)-y\left(\frac{x^{2}}{y}+3 \frac{y}{x}\right)}=\frac{\frac{2 x}{y}-\frac{3 y}{x^{2}}+\frac{6}{y^{2}}}{2 x^{2}+\frac{6 x}{y}-\frac{3 y^{2}}{x}}=\frac{\frac{2 x}{y}-\frac{3 y}{x^{2}}+\frac{6}{y^{2}}}{x y\left(\frac{2 x}{y}-\frac{3 y}{x^{2}}+\frac{6}{y^{2}}\right)}=\frac{1}{x y} $$
Let $xy=z$

Thus, we have an integrating factor
$$
\mu(x y)=\exp \left(\int \frac{1}{z} \mathrm{d} z\right)=e^{\log |z|}=z=x y
$$

Multiplying the original differential equation through by our integrating factor, we have
 $$\left(3 x^{2} y+6 x\right)+\left(x^{3}+3 y^{2}\right) \frac{d y}{d x}=0$$
We can see that this differential equation is exact because
$$
\frac{\partial}{\partial y}\left(3 x^{2} y+6 x\right)=3 x^{2}=\frac{\partial}{\partial x}\left(x^{3}+3 y^{2}\right)
$$
Thus, there exists a function $\psi(x, y)$ such that
\begin{equation}
\psi_{x}(x, y)=3 x^{2} y+6x
\tag{1}
\end{equation}
\begin{equation}
\psi_{y}(x, y)=x^{3}+3 y^{2}
\tag{2}
\end{equation}
Integrating (1) with respect to $x$, we get
$$
\psi(x, y)=x^{3} y+3 x^{2}+h(y)
$$
for some function $h$ of $y$. Next, differentiating with respect to $y,$ and equating
with ( 2)$,$ we get
$$
\psi_{y}(x, y)=x^{3}+h^{\prime}(y)
$$
Therefore,
$$
h^{\prime}(y)=3 y^{2}
$$
$$h(y)=y^{3}$$
and we have
$$
\psi(x, y)=x^{3} y+3 x^{2}+y^{3}
$$
Thus, the solutions of the differential equation are given implicitly by
$$
x^{3} y+3 x^{2}+y^{3}=C
$$

5
Quiz-1 / TUT0601 Quiz1
« on: September 27, 2019, 11:35:07 PM »
Solve the given differential equation.

The given differential equation is

$
\frac{d y}{d x}=\frac{x-e^{-x}}{y+e^{y}}
$

i.e. $\left(y+e^{y}\right) \frac{d y}{d x}=\left(x-e^{-x}\right)$

or $\left(e^{-x}-x\right)+\left(y+e^{y}\right) \frac{d y}{d x}=0$

This equation is of the form $M(x)+N(y) \frac{d y}{d x}=0$ and hence is separable

On rewriting this equation we have

$\left(y+e^{y}\right) d y=\left(x-e^{-x}\right) d x$

On integrating $\int\left(y+e^{y}\right) d y=\int\left(x-e^{-x}\right) d x$

i.e. $\frac{y^{2}}{2}+e^{y}=\frac{x^{2}}{2}+e^{-x}+c^{\prime}$
 
where $c'$ is an arbitrary constant of integration

i.e. $y^{2}+2 e^{y}=x^{2}+2 e^{-x}+2 c^{\prime}$

Put $2 c^{\prime}=c$, another arbitrary

i.e. $y^{2}-x^{2}+2\left(e^{y}-e^{-x}\right)=c$ is the required solution

where $y+e^{y} \neq 0$

Pages: [1]