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### Messages - Roro Sihui Yap

Pages: 1 [2]
16
##### Q1 / Re: Q1-P2
« on: September 29, 2016, 09:30:40 PM »
$u_{xyz} = \sin x + \sin y + \sin z$
$u_{yz} = - \cos x + x \sin y + x \sin z + f_{yz}(y,z)$
$u_{z} = - y \cos x - x \cos y + xy \sin z + f_z(y,z) + g_z(x,z)$
$u = - yz \cos x - xz \cos y - xy \cos z + f(y,z) + g(x,z) + h(x,y)$

17
##### Chapter 2 / Re: Solving the Hopf equation
« on: September 25, 2016, 10:06:30 PM »
Remember implicit function theorem? We need $F_u\ne 0$ where $F=F(x,t,u)=u-f(x,t,u)$. Since $F(x,0,u)=u)$ we have $F_u(x,0,u)=1>0$ and (\ref{A}) is fulfilled as long $F_u(x,t,u)>0$; so solution breaks when

F_u= 1-t f(x-ut)=0.

Let $s=x-ut$, then (\ref{A}) implies $u=f(s)$, (\ref{B}) implies $t=1/f(s)$, then $x=s+ut= s-f(s)/f'(s)$. So

\left\{\begin{aligned}
&x=s-f(s)/f'(s),\\
&t=f'(s)
\end{aligned}
\right.

Should the following be equation (2) instead ?
Since $F = u - f(x-ut)$
$F_u = 1 - fâ€™(x-ut)\cdot(-t) = 1+t f'(x-ut)=0$

Then, substitution of $s = x-ut$ implies $t = -1/f'(s)$
And subsequently $x = s + ut = s + f(s)\cdot\frac{-1}{f'(s)} = s - \frac{f(s)}{f'(s)}$

18
##### MT / Re: MT problem 5
« on: October 30, 2014, 09:50:59 AM »
I will in the future

19
##### MT / Re: MT problem 3
« on: October 30, 2014, 01:25:34 AM »
Let $t = ln x$
$x =e^t$

Let $\frac{dy}{dx} = e^{-t}\frac{dy}{dt}$
Let $\frac{d^2y}{dx^2} = e^{-2t}\frac{d^2y}{dt^2} - e^{-2t}\frac{dy}{dt}$
Let $\frac{d^3y}{dx^3} = e^{-3t}\frac{d^3y}{dt^3} - 3e^{-3t}\frac{d^2y}{dt^2} + 2e^{-3t}\frac{dy}{dt}$

$x^3 y''' + 6x^2 y'' + 5x y' -5y = x^2 \ln x \$
$e^{3t} (e^{-3t}\frac{d^3y}{dt^3} - 3e^{-3t}\frac{d^2y}{dt^2} + 2e^{-3t}\frac{dy}{dt}) + 6e^{2t} ( e^{-2t}\frac{d^2y}{dt^2} - e^{-2t}\frac{dy}{dt}) + 5e^t (e^{-t}\frac{dy}{dt}) - 5y = te^{2t}$
$\frac{d^3y}{dt^3} - 3\frac{d^2y}{dt^2} + 2\frac{dy}{dt}+ 6\frac{d^2y}{dt^2} - 6\frac{dy}{dt}+ 5\frac{dy}{dt} - 5y = te^{2t}$
$\frac{d^3y}{dt^3} + 3\frac{d^2y}{dt^2} + \frac{dy}{dt} - 5y = te^{2t}$
\begin{gather}  y''' + 3y'' + y' - 5y = te^{2t} \end{gather}

Find the solution to the homogeneous equation,
$r^3 + 3r^2 + r - 5 = 0$
$(r - 1)(r^2 + 4r + 5) = 0$
r = 1, -2 - i, -2 + i

The solution to the homogeneous equation is
$y = c_1e^t + c_2e^{-2t}cos(t) + c_3e^{-2t}sin(t)$
$y = c_1x + \frac{c_2}{x^2}cos(lnx) + \frac{c_3}{x^2}sin(lnx)$

Use equation 1 and the method of undetermined coefficients to find the non homogeneous solution
let $y_p = Ate^{2t} + Be^{2t}$
$y_p' = Ae^{2t} + 2Ate^{2t} + 2Be^{2t}$
$y_p'' = 4Ae^{2t} + 4Ate^{2t} + 4Be^{2t}$
$y_p''' = 12Ae^{2t} + 8Ate^{2t} + 8Be^{2t}$

Substitute into equation 1
$12Ae^{2t} + 8Ate^{2t} + 8Be^{2t} + 12Ae^{2t} + 12Ate^{2t} + 12Be^{2t} + Ae^{2t} + 2Ate^{2t} + 2Be^{2t} - 5Ate^{2t} - 5Be^{2t} = te^{2t}$
$8Ate^{2t} + 12Ate^{2t} + 2Ate^{2t} - 5Ate^{2t} = te^{2t}$
$17A = 1$
Therefore A = $\frac{1}{17}$
$12Ae^{2t} + 8Be^{2t} + 12Ae^{2t} + 12Be^{2t} + Ae^{2t} + 2Be^{2t} - 5Be^{2t} = 0$
$25A + 17B = 0$
Therefore B = $\frac{-25}{289}$

$y_p = \frac{1}{17}te^{2t} - \frac{25}{289}e^{2t}$
$y_p = \frac{x^2(ln x)}{17} - \frac{25x^2}{289}$

Therefore $y = c_1x + \frac{c_2}{x^2}cos(lnx) + \frac{c_3}{x^2}sin(lnx) + \frac{x^2(ln x)}{17} - \frac{25x^2}{289}$

20
##### MT / Re: MT problem 1
« on: October 30, 2014, 12:10:23 AM »
\begin{equation*}
\left(3x + \frac{6}{y}\right) + \left(\frac{x^2}{y}+\frac{3y}{x}\right)y'=0.
\end{equation*}
Let $M(x,y) = 3x + \frac{6}{y} \\$ and  $N(x,y) = \frac{x^2}{y}+\frac{3y}{x}$. Then  $M_y = \frac{-6}{y^2}$,  $N_x = \frac{2x}{y}+\frac{-3y}{x^2}$

The equation is not exact. Consider $\frac{M_y - N_x}{N(x,y)}$,
\begin{equation*}
\frac{M_y - N_x}{N(x,y)} = \frac{\frac{-6}{y^2} - \frac{2x}{y} + \frac{3y}{x^2}}{\frac{x^2}{y}+\frac{3y}{x}} = \frac{-6x^2 - 2x^3y + 3y^3}{x^4y + 3y^3x}
\end{equation*}
This is not a function of $x$ only. The integrating factor depending only on $x$ does not exist

Consider $\frac{N_x - M_y}{M(x,y)}$,
\begin{equation*}
\frac{N_x - M_y}{M(x,y)} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{3x + \frac{6}{y}} = \frac{6x^2 + 2x^3y - 3y^3}{3x^3y^2 + 6x^2y}
\end{equation*}
This is not a function of $y$ only. The integrating factor depending only on $y$ does not exist.

Consider $\frac{N_x - M_y}{(x)M(x,y) - (y)N(x,y)}$,
\begin{equation*}
\frac{N_x - M_y}{(x)M - (y)N} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{3x^2 + \frac{6x}{y} - x^2 - \frac{3y^2}{x}} = \frac{6x^2 + 2x^3y - 3y^3}{6x^3y + 2x^4y^2 - 3y^4x} = \frac{1}{xy}
\end{equation*}
This is a function of $xy$ only. $\frac{d\mu}{dxy} = \frac{\mu}{xy}$. Let $u = xy$;  $\frac{d\mu}{du} = \frac{\mu}{u}$,  $\mu = u$, $\mu = xy$

Multiply the equation by $xy$
\begin{equation*}
\left(3x^2y + 6x\right) + \left(x^3+3y^2\right)y'=0.
\end{equation*}
Let $M(x,y) = 3x^2y + 6x$ and  $N(x,y) = x^3+3y^2$.

There is a  $\Psi(x, y)$ such that:
\begin{gather}
\Psi _x(x, y) = M(x,y) = 3x^2y + 6x  \label{A}\\
\Psi _y (x, y) = N(x,y) = x^3+3y^2 \label{B}\end{gather}

Integrating equation (\ref{A}), $\Psi(x, y) = x^3y + 3x^2 + f(y)$, $\Psi_y (x, y) = x^3 + f'(y)$.

Comparing with equation (\ref{B}), $f'(y) = 3y^2$ and $f(y) = y^3$

Therefore $\Psi(x, y) = x^3y + 3x^2 + y^3 = c$.
When $x=1$, $y=2$,   $2 + 3 + 8 = c$, $c = 13\implies \Psi(x, y) = x^3y + 3x^2 + y^3 = 13$

21
##### MT / Re: MT problem 2
« on: October 29, 2014, 11:32:16 PM »
First find the solution to the homogeneous equation.

$r^2 + 2r + 1 = 0 \\$
$(r + 1)^2 = 0 \\$
$r = -1, -1 \\$

The solution to the homogeneous equation is
$z = c_1e^{-x} + c_2xe^{-x}$

Use the method of variation of parameters
let $z_p = u_1e^{-x} + u_2xe^{-x}$

\begin{gather}
u_1'e^{-x} + u_2'xe^{-x} = 0,\label{eq-1} \\
-u_1'e^{-x} + u_2'(e^{-x}-xe^{-x}) = e^{-x}\ln x  \label{eq-2} \\
\end{gather}

Add equation (\ref{eq-1}) and (\ref{eq-2})  together,
$u_2'e^{-x} = e^{-x}\ln x \\$
$u_2' = \ln x$
$u_2 = x\ln x - x$

From equation 1,
$u_1' = -xu_2'$
$u_1' = -x\ln x$
$u_1 = x^2/4 - x^2\ln x/2$

$z_p = x^2e^{-x}/4 - x^2e^{-x}\ln x/2 + x^2e^{-x}\ln x - x^2e^{-x} \\$
$z_p = -3x^2e^{-x}/4 + x^2e^{-x}\ln x/2$

The solution is $z = c_1e^{-x} + c_2xe^{-x} -3x^2e^{-x}/4 + x^2e^{-x}\ln x/2$
When x = 1, z = -2
$-2 = c_1e^{-1} + c_2e^{-1} - (3/4)e^{-1} \\$
$-2e + 3/4 = c_1 + c_2$
let $c_1 = -2e$ and $c_2 = 3/4$

The solution satisfying z(1) = -2 is
$z = -2e^{1-x} + 3xe^{-x}/4 -3x^2e^{-x}/4 + \frac{1}{2}x^2e^{-x}\ln x$

22
##### MT / Re: MT problem 5
« on: October 29, 2014, 11:07:37 PM »
First, find the eigenvalues of the matrix A.
\begin{equation*} A = \begin{bmatrix}
1 & 1 & 2
\\1 & 2 & 1
\\ 2 & 1 & 1\end{bmatrix} \end{equation*}

\begin{equation*}
\left| \begin{matrix} 1-\lambda & 1 & 2 \\1 &  2-\lambda & 1 \\ 2& 1 & 1-\lambda\end{matrix}\right| = 0  \notag
\end{equation*}

\begin{gather}
(1 - \lambda)( (2-\lambda)(1-\lambda) - 1) - (1 - \lambda - 2 ) + 2 (1 - 4 + 2\lambda) = 0,\label{eq-1} \\
(1 - \lambda)(\lambda^2 - 3\lambda + 1) + (1 + \lambda) +  (-6 + 4\lambda) = 0 \label{eq-2} \\
-\lambda^3 + 4\lambda^2 + \lambda - 4 = 0 \label{eq-3} \\
-(\lambda - 4)(\lambda - 1)(\lambda + 1) = 0 \label{eq-4} \\
\end{gather}
The eigenvalues are $\lambda = 1, \lambda = -1,\ \lambda = 4$.

When $\lambda = 1$,
\begin{equation*}
A-\lambda I= \begin{bmatrix}
0 & 1 & 2
\\1 & 1 & 1
\\ 2 & 1 & 0\end{bmatrix} \cong
\begin{bmatrix}
1 & 0 & -1
\\0 & 1 & 2
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(1, -2, 1)^T$ and corresponding solution is  $x^{(1)} = c_1e^t(1, -2, 1)^T$.

When $\lambda = -1$,
\begin{equation*}
A-\lambda I= \begin{bmatrix}
2 & 1 & 2
\\1 & 3 & 1
\\ 2 & 1 & 2\end{bmatrix} \cong
\begin{bmatrix}
1 & 0 & 1
\\0 & 1 & 0
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(1, 0, -1)^T$ and corresponding solution is  $x^{(2)} = c_2e^{-t}(1, 0, -1)^T$.

When $\lambda = 4$,
\begin{equation*}
A-\lambda I= \begin{bmatrix}
-3 & 1 & 2
\\1 & -2 & 1
\\ 2 & 1 & -3\end{bmatrix} \cong
\begin{bmatrix}
1 & 0 & -1
\\0 & 1 & -1
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(1, 1, 1)^T$ and corresponding solution is  $x^{(3)} = c_3e^{4t}(1, 1, 1)^T$.

The general solution is $(x, y, z)^T = c_1e^t(1, -2, 1)^T + c_2e^{-t}(1, 0, -1)^T + c_3e^{4t}(1, 1, 1)^T$

23
##### Quiz 3 / Re: Q3 problem 2 (night sections)
« on: October 25, 2014, 08:35:32 PM »
First, find the eigenvalues of the matrix A.

\begin{equation*}
\left| \begin{matrix} 1-\lambda & 0 & 0 \\2 &  1-\lambda & -2 \\ 3& 2 & 1-\lambda\end{matrix}\right| = 0  \notag
\end{equation*}
\begin{gather}
(1 - \lambda)( (1-\lambda)(1-\lambda) + 4) = 0,\label{eq-1} \\
(1 - \lambda)( \lambda^2 - 2\lambda + 5) = 0. \label{eq-2}
\end{gather}
The eigenvalues are $\lambda = 1, \lambda = 1 + 2i,\ \lambda = 1- 2i$.

When $\lambda = 1$,
\begin{equation*}
A-\lambda I= \begin{bmatrix}
0 & 0 & 0
\\2 & 0 & -2
\\ 3 & 2 & 0\end{bmatrix} \cong
\begin{bmatrix}
1 & 0 & -1
\\0 & 2 & 3
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(2, -3, 2)^T$ and corresponding solution is  $x^{(1)} = c_1e^t(2, -3, 2)^T$.

When $\lambda = 1 + 2i$,
\begin{equation*}
A-\lambda I=\begin{bmatrix}
-2i & 0 & 0
\\2 & -2i & -2
\\ 3 & 2 & -2i\end{bmatrix}\cong
\begin{bmatrix}
1 & 0 & 0
\\0 & i & 1
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(0, i, 1) ^T$ and corresponding solution is
\begin{equation*}
(0, i, 1)^T e^{(1+2i)t} = (0, i, 1)^T (e^t)( e^{2it})  = (0, i, 1)^T (e^t)(\cos  2t + i\sin 2t) = (0, i\cos 2t - \sin  2t, \cos 2t + i\sin 2t)^T (e^t) $=(e^t)[ (0, -\sin 2t, \cos 2t)^T +i(0, \cos 2t, \sin 2t)^T] \end{equation*} The general solution is$x = e^t\left[ c_1 (2, -3, 2)^T + c_2 (0, -\sin  2t, \cos  2t)^T +c_3 (0, \cos  2t, \sin 2t)^T\right]$24 ##### TT1 / Re: TT1-problem 2 « on: October 09, 2014, 12:32:32 PM » 2b$y_1(x)=x$,$y_1'(x)=1$,$y_1''(x)=0$. Substituting into the equation we get, x^3(\ln x+1)\cdot (0) -(2\ln x+3)x^2 + (2\ln x+3) x^2 = 0$y_1(x)=x$is indeed a solution. We know that W = c(x^2 (\ln(x) +1)) and also$W = \det \begin{bmatrix} x & g(t) \\1 &  gâ€™(t)\end{bmatrix}$or simply$W = \left| \begin{matrix} x & g(t) \\1 &  gâ€™(t)\end{matrix}\right|$So, xg'(t) - g(t) = c(x^2 (ln(x) +1))\\ g'(t) - g(t)/x = c(xln(x) +x) Find an integrating factor$d\mu/dx = e^{-\int1/x} $.$ \mu = x^{-1} $. Multiply by$ x^{-1}$. Throughout the equation$g'(t)/x - g(t)/x^2 = c(\ln(x) + 1)$,$[g(t)/x]' = c\ln x + c$,$ g(t)/x = cx\ln x - cx + cx \implies  g(t) = cx^2\ln x $-- The second solution . 25 ##### TT1 / Re: TT1-problem 2 « on: October 09, 2014, 12:28:39 PM » \begin{equation*} y''(x) - \frac{(2\ln x+3)}{x(\ln x+1)} y'(x) + \frac{(2\ln x+3)}{x^2(\ln x+1)}y(x)=0 \end{equation*} Using Abel's Theorem, W=ce^{\int{\frac{(2\ln x+3)}{x(\ln x+1)}}}dt \int{\frac{(2\ln x+3)}{x(\ln x+1)}dt} = \int{\frac{(2\ln x+4)}{x(\ln x+1)}dt} - \int{\frac{1}{x(\ln x+1)}dt} \\ \int{\frac{(2\ln x+3)}{x(\ln x+1)}dt} = 2ln(xln(x) + x) - ln(ln(x) +1) \\ \int{\frac{(2\ln x+3)}{x(\ln x+1)}dt} = ln(x^2 (lnx +1)) Therefore, W = c(x^2 (ln(x) +1)) 26 ##### TT1 / Re: TT1-problem 1 « on: October 09, 2014, 12:09:00 PM » \begin{equation*} y + (2xy - e^{-2y})y' = 0 \ . \end{equation*} Let$M(x,y) = y$,$N(x,y) =2xy - e^{-2y}$. Then$M_y(x,y) = 1$,$N_x(x,y) = 2y$. The equation is not exact. Lets find an integrating factor to make it exact.$d\mu/dy = (N_x  - M_y)\mu / M\implies
d\mu/dy = (2y - 1)\mu / y \implies
d\mu/ \mu  = (2 - (1/y))dy\implies
\ln \mu = 2y - lny \implies
\mu = e^{2y - lny} \implies
\mu = e^{2y} / y $Now multiply the equation by$\mu = e^{2y} / y$e^{2y} + (2xe^{2y} - 1/y ) yâ€²= 0 Now$M(x,y) = e^{2y}$,$N(x,y) =(2xe^{2y} - 1/y )$. There is a$\Psi(x, y)$such that: \begin{gather} \Psi _x(x, y) = M(x,y) = e^{2y}\\ \Psi _y (x, y) = N(x,y) = (2xe^{2y} - 1/y ) \end{gather} Integrating (2) we have$\Psi (x, y) = xe^{2y} + f(y)$. Using this, differentiate to get$\Psi_y (x, y) = 2xe^{2y} + f'(y) $. Compare this with (3):$f'(y) = -1/y$.$Meaning f(y) = -ln|y|$So, \Psi (x, y) = xe^{2y} - ln|y| = C When x = 1, y = -2$e^{-4} - ln |-2| = C $The solution is xe^{2y} - ln|y| = e^{-4} - ln 2 27 ##### Quiz 2 / Re: Quiz 2 Problem 1 (night sections) « on: October 02, 2014, 11:16:18 AM » Consider Wronskian$W = \det \begin{bmatrix} f(t) \ \ \ g(t) \\fâ€™(t) \ \ gâ€™(t)\end{bmatrix}$3e^{4t} = f(t)gâ€™(t) - g(t)fâ€™(t) We know$f(t) = e^{2t}$. Therefore$fâ€™(t) = 2e^{2t}$. Substitute the terms in $$3e^{4t} = (e^{2t})gâ€™(t) - (2e^{2t})g(t)$$ Divide all terms by$(e^{2t})$\begin{gather} 3e^{2t} = gâ€™(t) - 2g(t) \\ gâ€™(t) - 2g(t) = 3e^{2t} \end{gather} We need to find an integrating factor$\mu (t) = e^{\int -2\,dt}= e^{-2t}$. Multiply all terms by$e^{-2t}$\begin{gather} gâ€™(t)e^{-2t} - 2g(t)e^{-2t} = 3 \\ [g(t)e^{-2t}]' = 3 \\ g(t)e^{-2t} = 3t + c \\ g(t) = 3te^{2t} + ce^{2t} \end{gather} Niceâ€”I made minor improvements. V.I. 28 ##### Quiz 1 / Re: Q1 problem 1 (L5101) « on: September 25, 2014, 12:31:06 AM » (3 x^2 y + 2xy + y^3) + (x^2 + y^2) yâ€²= 0\label{A} Let$M(x,y) = 3(x^2)y + 2xy + y^3$,$N(x,y) = x^2 + y^2$. Then$M_y(x,y) = 3x^2 + 2x + 3y^2$,$N_x(x,y) = 2x$. Equation (\ref{A}) is not exact. Lets try to find an integrating factor$\mu=\mu(x)$to make it exact.$d\mu/dx = (M_y  - N_x)\mu / N\implies
d\mu/dx = (3x^2 + 2x + 3y^2 - 2x)\mu / N \implies
d\mu/dx = 3(x^2 + y^2) \mu / (x^2 + y^2)\implies
d\mu/dx = 3Î¼\implies
d\mu/ \mu  = 3 dx\implies
\ln \mu = 3x \implies
\mu = e^{3x} $Now multiply the equation(\ref{A}) by$\mu = e^{3x}$\bigl((3 x^2y + 2xy +y^3)e^{3x}\bigr)+ \bigl((x^2+y^2)e^{3x}\bigr)yâ€²= 0 \label{B} Now$M(x,y) = 3(x^2y + 2xy +y^3)e^{3x}$,$N(x,y) =(x^2+y^2)e^{3x}$. Then$M_y(x,y) = (3 x^2  + 2x  + 3y^2)e^{3x}$,$N_x(x,y) = (2x3+x^2) + 3y^2)e^{3x}$,$M_y(x,y) = N_x(x,y)$. Therefore the equation is exact. No need to check: it is exact due to construction of$\mu$. V.I. There is a$\Psi(x, y)$such that: \begin{gather} \Psi _x(x, y) = M(x,y) =3(x^2y + 2xy +y^3)e^{3x},\label{C}\\ \Psi _y (x, y) = N(x,y) = (x^2+y^2)e^{3x}.\label{D} \end{gather} Integrating (\ref{C}) we have$\Psi (x, y) = (x^2+ \frac{1}{3}  y^3)e^{3x} + f(y)$. Using this, differentiate to get$\Psi_y (x, y) =(x^2+y^2)e^(3x) + f'(y) $. Easier to start from (\ref{D}) V.I. Compare this with (\ref{D}):$f'(y) = 0$.$Meaning f(y) = C$, where$C$is some constant So, \Psi (x, y) = (x^2+ \frac{1}{3} y^3)e^{3x} +C=0$  is a solution.

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