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.


Messages - EroSkulled

Pages: [1]
1
Quiz-5 / TUT 0401 Quiz 5
« on: November 03, 2019, 06:19:39 PM »
Verify that the given functions of y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
$(1-t)y''+ty'-y=2(t-1)^{2}e^{-t},0<t<1$
$y_{1}(t)=e^{t}, y_{2}(t)=t$
Hence,
$$\begin{cases}
y_{1}(t)=e^t\\
y_{1}'(t)=e^t\\
y_{1}''(t)=e^t\\
\end{cases} ,  \begin{cases}
y_{2}(t)=t\\
y_{2}'(t)=1\\
y_{2}''(t)=0\\
\end{cases}
$$
Subsitute back into the homogeneous equation:
$$(1-t)y''+ty'-y=0$$
\begin{equation}
(1-t)e^t+te^t-e^t=0
\end{equation}
\begin{equation}
e^t(1-t+t-1)=0
\end{equation}
\begin{equation}
0=0
\end{equation}
Verified $y_1$ satisfy the corresponding homogeneous equation.
\begin{equation}
(1-t)0+t*1-t=0
\end{equation}
\begin{equation}
t-t=0
\end{equation}
\begin{equation}
0=0
\end{equation}
Verified $y_2$ satisfy the corresponding homogeneous equation.
Now we find particular solution of the given nonhomogeneous equation by first dividing it by $1-t$:
\begin{equation}
y''+\frac{t}{1-t}y'-\frac{1}{1-t}y=2(1-t)e^{-t}
\end{equation}
Then
\begin{equation}
g(t)=2(1-t)e^{-t}
\end{equation}
\begin{equation}
W[y_1,y_2]=\begin{vmatrix}
{y_{1}(t)}&{y_{2}(t)}\\
{y_{1}'(t)}&{y_{2}'(t)}\\
\end{vmatrix}=(1-t)e^t
\end{equation}
\begin{equation}
W_1[y_1,y_2]=\begin{vmatrix}
{0}&{y_{2}(t)}\\
{1}&{y_{2}'(t)}\\
\end{vmatrix}=-t
\end{equation}
\begin{equation}
W_2[y_1,y_2]=\begin{vmatrix}
{y_{1}(t)}&{0}\\
{y_{1}'(t)}&{1}\\
\end{vmatrix}=e^t
\end{equation}
Then:
\begin{equation}
Y(t)=y_{1}\int{\frac{W_1[y_1,y_2]*g(t)}{W[y_1,y_2]}dt}+y_{2}\int{\frac{W_2[y_1,y_2]*g(t)}{W[y_1,y_2]}dt}
\end{equation}
\begin{equation}
Y(t)=e^t\int{\frac{-t*2(1-t)e^{-t}}{(1-t)e^t}dt}+t\int{\frac{e^t*2(1-t)e^{-t}}{(1-t)e^t}dt}
\end{equation}
\begin{equation}
Y(t)=-2e^t\int{\frac{t}{e^{2t}}dt}+2t\int{\frac{1}{e^t}dt}
\end{equation}
\begin{equation}
Y(t)=-2e^t((-\frac{t}{2}-\frac{1}{4})e^{-2t})+2t(-2e^{-t})
\end{equation}
\begin{equation}
Y(t)=(\frac{1}{2}-t)e^{-t}
\end{equation}
Therefore the particular solution of the given nonhomogeneous equation is
$(\frac{1}{2}-t)e^{-t}$
(By setting constants to 0)



2
Term Test 1 / Re: Problem 1 (morning)
« on: October 23, 2019, 07:30:45 AM »
Quote from: rj127 on October 23, 2019, 06:37:18 AM
(a) Let $$M=-ySin(x)+y^3Cos(x)$$
$$N=3Cos(x)+5y^2Sin(x)$$
Then$$M_y=-Sin(x)+3y^2Cos(x)$$
$$N_x=-3Sinx(x)+5y^2Cos(x)$$
$$R=\frac{M_y-N_x}{M}=\frac{2Sin(x)-2Cos(x)}{-ySin(x)+y^3Cos(x)}=\frac{2(Sin(x)-y^2Cos(x))}{-y(Sin(x)-y^2Cos(x))}=-\frac{2}{y}$$
$$\mu=e^{-\int_Rdy}=e^{\int_\frac{2}{y}dy}=e^{2lny}=e^ln(y^2)=y^2$$

Multiple both side by $$\mu=y^2$$
$$y^2(-ySin(x)+y^3Cos(x))+y^2(3Cos(x)+5y^2Sin(x))=0$$
$$M'=-y^3Sin(x)+y^5Cos(x)$$,$$N'=3y^2Cos(x)+5y^4Sin(x)$$
$$\exists\phi_x,y,such that \phi_x=M',\phi_y=N$$
$$\phi=\int_M'dx=\int_-y^3Sin(x)=y^3Cos(x)+y^5Sin(x)+h(y)$$
$$\phi_y=3y^2Cos(x)+5y^4Sin(x)+h'(y)=N'$$
$$h'(y)=0$$
$$h(y)=c$$

$$\therefore \phi=y^3Cos(x)+y^5Sin(x)=c$$
(b)When $$y(\frac{\pi}{4})=\sqrt{2}$$
$$(\sqrt{2})^3Cos(\frac{\pi}{4})+(\sqrt{2}^5)Sin(\frac{\pi}{4})=2\sqrt{2}*\frac{1}{\sqrt{2}}+(4\sqrt{2}*\frac{1}{\sqrt{2}})=2+4=6$$
$$\therefore c=6$$

$$\phi=y^3Cos(x)+y^5Sin(x)=6$$
Above solution is not typed well in correct format so I posted mine as well.

3
Term Test 1 / Re: Problem 1 (morning)
« on: October 23, 2019, 07:27:33 AM »
Solve :$(-y\sin(x)+y^{3}\cos(x))+(3\cos(x)+5y^{2}\sin(x))y'=0$
\begin{equation}
M=-y\sin(x)+y^{3}\cos(x), N=3\cos(x)+5y^{2}\sin(x)
\end{equation}
\begin{equation}
M_y=-\sin(x)+3y^{2}\cos(x), N_x=-3\sin(x)+5y^{2}\cos(x)
\end{equation}
\begin{equation}
R_1=\frac{N_x-M_y}{M}=\frac{-3\sin(x)+5y^{2}\cos(x)+\sin(x)-3y^{2}\cos(x)}{-y\sin(x)+y^{3}\cos(x)}=\frac{-2\sin(x)+2y^{2}\cos(x)}{y(-\sin(x)+y^{2}\cos(x))}=\frac{2}{y}
\end{equation}
\begin{equation}
\mu=e^{\int{R_1}{dy}}=e^{\int{\frac{2}{y}}{dy}}=e^{2\ln{y}}=y^2
\end{equation}
We then multiply both side of the original equation by $y^2$ so that it will become EXACT and hence we can continue to find $\phi(x,y)$
\begin{equation}
(-y^{3}\sin(x)+y^{5}\cos(x))+(3y^{2}\cos(x)+5y^{4}\sin(x))y'=0
\end{equation}
\begin{equation}
\phi(x,y)=\int{-y^{3}\sin(x)+y^{5}\cos(x)}{dx}=y^{3}\cos(x)+y^{5}\sin(x)+h(y)
\end{equation}
\begin{equation}
\phi(x,y)_y=3y^{2}\cos(x)+5y^{4}\sin(x)+h'(y)\cong 3y^{2}\cos(x)+5y^{4}\sin(x)
\end{equation}
Hence we know $h'(y)=0$
Then $h(y)=C$
\begin{equation}
\phi(x,y): y^{3}\cos(x)+y^{5}\sin(x)=C
\end{equation}
Initial Value: $y(\frac{\pi}{4})=\sqrt{2}$
Plug in equation above, we get the following:
\begin{equation}
(\sqrt{2})^{3}\cos(\frac{\pi}{4})+(\sqrt{2})^{5}\sin(\frac{\pi}{4})=C
\end{equation}
\begin{equation}
2\sqrt{2}\frac{1}{\sqrt{2}}+4\sqrt{2}\frac{1}{\sqrt{2}}=C
\end{equation}
\begin{equation}
C=6
\end{equation}
We get solution:
\begin{equation}
y^{3}\cos(x)+y^{5}\sin(x)=6
\end{equation}

No post after this is needed. V.I.
Instead of sequence single equations it would be better to use multiline environment like gather or gather* to avoid excessive vertical spacing
Code: [Select]
\begin{gather} EQUATION \\ EQUATION \\  .... \end{gather} If there is no text between them, as MathJax does not support \intertext{  } LaTeX command

4
Quiz-3 / TUT0401 QUIZ3
« on: October 11, 2019, 02:00:00 PM »
Solve $y''+5y'+3y=0$ with initial condition $y(0)=1, y'(0)=0$

Characteristic equation: $r^2+5r+3=0$
Then we get two solutions:$r_1=\frac{-5+\sqrt{13}}{2}$, $r_2=\frac{-5-\sqrt{13}}{2}$
Hence general solution:
\begin{equation}
y(t)=Ae^{t\frac{-5+\sqrt{13}}{2}}+Be^{t\frac{-5-\sqrt{13}}{2}}
\end{equation}

\begin{equation}
y’(t)= \frac{-5+\sqrt{13}}{2} Ae^{t\frac{-5+\sqrt{13}}{2}}-\frac{5+\sqrt{13}}{2} Be^{t\frac{-5-\sqrt{13}}{2}}
\end{equation}

Since $y(0)=1, y'(0)=0$

Solve equation sets above for A and B, we get $A=\frac{5+\sqrt{13}}{2\sqrt{13}}$, $B=\frac{-5+\sqrt{13}}{2\sqrt{13}}$

Hence the solution of equation is:

\begin{equation}

y(t)= \frac{5+\sqrt{13}}{2\sqrt{13}} e^{t\frac{-5+\sqrt{13}}{2}}+ \frac{-5+\sqrt{13}}{2\sqrt{13}} e^{t\frac{-5-\sqrt{13}}{2}}

\end{equation}

5
Quiz-2 / TUT0401 quiz2 solution
« on: October 04, 2019, 02:00:01 PM »
Solve$ \frac{x}{(x^2+y^2)^\frac{3}{2}} + \frac{y}{(x^2+y^2)^\frac{3}{2}}\frac{dy}{dx}=0$

Let $M=\frac{x}{(x^2+y^2)^\frac{3}{2}}$ and $N=\frac{y}{(x^2+y^2)^\frac{3}{2}}$
We want to find $N_x$ and $M_y$
\begin{equation}
  M_y=\frac{d}{dy}\frac{x}{(x^2+y^2)^\frac{3}{2}}=-\frac{3xy}{(x^2+y^2)^\frac{5}{2}}
\end{equation}
\begin{equation}
  N_x=\frac{d}{dx}\frac{y}{(x^2+y^2)^\frac{3}{2}}=-\frac{3xy}{(x^2+y^2)^\frac{5}{2}}
\end{equation}
\begin{equation}
  N_x=M_y
\end{equation}
We conclude the given equation is exact. We integrate to find $\varphi(x,y)$
Note $\varphi_x(x,y)=M$, integrate both side, we get:
\begin{equation}
  \varphi(x,y)=\int{Mdx}=\int{\frac{x}{(x^2+y^2)^\frac{3}{2}}dx}
\end{equation}
Using u subsitution with $u=x^2+y^2$, we get:
\begin{equation}
  \varphi(x,y)=\frac{1}{2}\int{\frac{1}{u^\frac{3}{2}}du}=-u^{-\frac{1}{2}}=-(x^2+y^2)^{-\frac{1}{2}}+h(y)
\end{equation}
Note $\varphi_y(x,y)\equiv N$, then:
\begin{equation}
  \varphi_y(x,y)=\frac{d}{dy}(-(x^2+y^2)^{-\frac{1}{2}}+h(y))
\end{equation}
\begin{equation}
  \varphi_y(x,y)=\frac{y}{(x^2+y^2)^\frac{3}{2}}+h'(y)\equiv \frac{y}{(x^2+y^2)^\frac{3}{2}}
\end{equation}
Hence we can conlude $h'(y)=0$
Then $h(y)=C$
\begin{equation}
  \varphi(x,y)=-(x^2+y^2)^{-\frac{1}{2}}+C
\end{equation}
The implicit solutoin is $-(x^2+y^2)^{-\frac{1}{2}}=C$

Pages: [1]