Author Topic: HA1 problem 3  (Read 897 times)

Victor Ivrii

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HA1 problem 3
« on: January 20, 2015, 06:49:21 AM »
Solutions to be posted as a "Reply" only after January 22, 21:00

Find the solution of
\begin{equation}
\left\{\begin{aligned}
&u_x+3u_y=xy,\\
 &u|_{x=0}=0.
\end{aligned} \right.\label{eq-HA1.4}
\end{equation}

Yiyun Liu

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Re: HA1 problem 3
« Reply #1 on: January 22, 2015, 09:06:03 PM »
question 3

Biao Zhang

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Re: HA1 problem 3
« Reply #2 on: January 22, 2015, 10:27:42 PM »
HA1-3

Ping Wei

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Re: HA1 problem 3
« Reply #3 on: January 23, 2015, 10:33:14 AM »
By examining Integral Lines: $\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}$; then we got $3x=y+C$ where $C$ is some constant.
$y=3x−C$

Then again from the Integral Lines:

dx(xy)=du

dx(x(3x−C))=du

u=x^3−C/2x^2+C1

Then by the initial condition:

u(x=0)=0

C1=0

Therefore,C=3x−y

Therefore,

u(x,y)=x^3−1/2x^2C

$u(x,y)=x^3−\frac{1}{2}x^2(3x−y)$

$u(x,y)=−\frac{1}{2}x^3+\frac{1}{2}x^2y$
« Last Edit: January 23, 2015, 11:07:47 AM by Victor Ivrii »

Victor Ivrii

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Re: HA1 problem 3
« Reply #4 on: January 23, 2015, 11:04:55 AM »
Nice

Ping, please fix your post (modify, look what I did with few  first and last lines!)
« Last Edit: January 23, 2015, 11:08:33 AM by Victor Ivrii »

Mark Nunez

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Re: HA1 problem 3
« Reply #5 on: January 23, 2015, 04:07:40 PM »
Still learning how to use LaTeX from MathType. :-\

Victor Ivrii

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Re: HA1 problem 3
« Reply #6 on: January 23, 2015, 04:11:47 PM »
Mark, typing rather than posting typed solutions convered to png is preferred.

Actually I have seen LaTeX codes from MathType and they are something you should not learn or try to imitate (overly complicated very custom code. Something completely opposite to good practice)

Also we seem to finish with HA1 but Web Bonus problems are untouched!

Yiyun Liu

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Re: HA1 problem 3
« Reply #7 on: January 28, 2015, 06:26:39 PM »
<sol>:The equation is linear,thus,the characteristic equation is below:$$\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac { dx }{ 1 } =\frac { dy }{ 3 } =\frac { du }{ xy } \\ \quad \quad \quad \quad \quad Solve\quad theO.D.E:\\ \quad \quad \quad \quad \quad \quad \quad \quad y=3x+C\\ \quad \quad \quad \quad \quad \quad \quad \quad (xy)dx=du\quad →\quad dx(x(3x+C))=du\\ \quad \quad \quad \quad \quad hence,u=x^{ 3 }+\frac { c{ x }^{ 2 } }{ 2 } +K,\quad whereC,\quad K\quad are\quad constants.\\ \quad \quad \quad \quad \quad solve\quad IVP:\\ \quad \quad \quad \quad \quad \quad \quad \quad K=0\\ \quad \quad \quad \quad \quad \quad \quad \quad C=-3x+y\\ \quad \quad \quad \quad \quad \quad \quad \quad u=x^{ 3 }+x^{ 2 }\frac { (-3x+y) }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =-\frac { { x }^{ 3 } }{ 2 } +\frac { y{ x }^{ 2 } }{ 2 } \\ \quad \quad \quad \quad check:\\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ x }=-\frac { 3{ x }^{ 2 } }{ 2 } +xy\\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ y }=\frac { { x }^{ 2 } }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ x }+3u_{ y }=xy\\ \\  $$
I just try whether what I typed could be shown

Yes--but you need to surround it by double dollars (I did it for you) . Still this is extremely bad code with all these \quad and many {} are completely unnecessary; also long pieces of text (like the whole line) should not be in formulas and shorter ones should be tagged as \text{…}. And better to break the source into lines (in the logical places, where \\ are)--V.I.
« Last Edit: January 28, 2015, 07:30:10 PM by Victor Ivrii »