Toronto Math Forum

APM346-2015S => APM346--Home Assignments => HA1 => Topic started by: Victor Ivrii on January 20, 2015, 06:49:21 AM

Title: HA1 problem 3
Post by: Victor Ivrii 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}

Title: Re: HA1 problem 3
Post by: Yiyun Liu on January 22, 2015, 09:06:03 PM

question 3

Title: Re: HA1 problem 3
Post by: Biao Zhang on January 22, 2015, 10:27:42 PM

HA1-3

Title: Re: HA1 problem 3
Post by: Ping Wei 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

$u(x,y)=x^3−\frac{1}{2}x^2(3x−y)$

$u(x,y)=−\frac{1}{2}x^3+\frac{1}{2}x^2y$

Title: Re: HA1 problem 3
Post by: Victor Ivrii 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!)

Title: Re: HA1 problem 3
Post by: Mark Nunez on January 23, 2015, 04:07:40 PM

Still learning how to use LaTeX from MathType. :-\

Title: Re: HA1 problem 3
Post by: Victor Ivrii 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!

Title: Re: HA1 problem 3
Post by: Yiyun Liu 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.