DE Marathon

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June 14th 2008, 04:16 PM
Chris L T521

DE Marathon

Mathstud's Limit Marathon has inspired me to start a Differential Equation Marathon. It will have similar rules : "after someone completes a DE and gets the OK from the DE's poster, he/she posts their own DE."

There are no restrictions on how you may solve them. But please show all the steps that led you to your solution.

Here's an easy one:

Solve $\bigg(e^x\sin(y)+\tan(y)\bigg)\,dx+\bigg(e^x\cos(y )+x\sec^2(y)\bigg)\,dy=0$
June 14th 2008, 04:56 PM
Chris L T521

HINT : It's an Exact Equation...
June 14th 2008, 05:00 PM
Krizalid

Quote:

Originally Posted by Chris L T521

HINT : It's an Exact Equation...

Easy Chris, there's no rush on it. Give a Hint when the problem has more hours without a solution.
June 15th 2008, 04:05 AM
galactus

$(e^{x}sin(y)+tan(y))dx+(e^{x}cos(y)+xsec^{2}(y))dy =0$

$\frac{{\partial}M}{{\partial}y}=\frac{e^{x}cos^{3} (y)+1}{cos^{2}(y)}=\frac{{\partial}N}{{\partial}x}$

$M(x,y)=\frac{{\partial}f}{{\partial}x} \;\ and \;\ N(x,y)=\frac{{\partial}f}{{\partial}y}$

Assume $\frac{{\partial}f}{{\partial}y}=N(x,y)$

$\frac{{\partial}f}{{\partial}y}=e^{x}cos(y)+xsec^{ 2}(y)$

$f(x,y)=\int{e^{x}cos(y)}dy+\int{xsec^{2}(y)}dy$

$f(x,y)=e^{x}sin(y)+xtan(y)+h(x)$

$\frac{{\partial}f}{{\partial}x}=(e^{x}cos(y)+1)tan (y)+h'(x)$

$\frac{{\partial}f}{{\partial}x}=(e^{x}cos(y)+1)tan (y)+h'(x)=e^{x}sin(y)+tan(y)$

$h'(x)=0$

Therefore, a family of solutions is

$\boxed{f(x,y)=e^{x}sin(y)+xtan(y)+C=0}$

May be a typo somewhere. Give me the go ahead and i will post one. Are word problems OK?. That is, something like Newton's cooling as an example.
June 15th 2008, 04:18 PM
Chris L T521

Quote:

Originally Posted by galactus

$(e^{x}sin(y)+tan(y))dx+(e^{x}cos(y)+xsec^{2}(y))dy =0$

$\frac{{\partial}M}{{\partial}y}=\frac{e^{x}cos^{3} (y)+1}{cos^{2}(y)}=\frac{{\partial}N}{{\partial}x}$

$M(x,y)=\frac{{\partial}f}{{\partial}x} \;\ and \;\ N(x,y)=\frac{{\partial}f}{{\partial}y}$

Assume $\frac{{\partial}f}{{\partial}y}=N(x,y)$

$\frac{{\partial}f}{{\partial}y}=e^{x}cos(y)+xsec^{ 2}(y)$

$f(x,y)=\int{e^{x}cos(y)}dy+\int{xsec^{2}(y)}dy$

$f(x,y)=e^{x}sin(y)+xtan(y)+h(x)$

$\frac{{\partial}f}{{\partial}x}=(e^{x}cos(y)+1)tan (y)+h'(x)$

$\frac{{\partial}f}{{\partial}x}=(e^{x}cos(y)+1)tan (y)+h'(x)=e^{x}sin(y)+tan(y)$

$h'(x)=0$

Therefore, a family of solutions is

$\boxed{f(x,y)=e^{x}sin(y)+xtan(y)+C=0}$

May be a typo somewhere. Give me the go ahead and i will post one. Are word problems OK?. That is, something like Newton's cooling as an example.

That's good! However, I was taught that the solution should have the form $f(x,y)=C$ , so my answer was $e^{x}\sin(y)+x\tan(y)=C$ .

Word problems are fine. They can be challenging (depending on how you interpret the question to write your DE); hence, that's the reason why I like them. You can post your question now, Galactus.
June 15th 2008, 05:30 PM
Mathstud28

No one is posting any, but here is one I saw before.

$y''+2y'+y=e^{-x}\ln(x)$
June 15th 2008, 05:51 PM
Chris L T521

Quote:

Originally Posted by Mathstud28

No one is posting any, but here is one I saw before.

$y''+2y'+y=e^{-x}\ln(x)$

Find the complementary solution:

$y''+2y'+y=0\implies r^2+2r+1=0\implies r=-1$ with multiplicity 2.

Thus, $y_c=c_1e^{-x}+c_2xe^{-x}$

Apply the technique of variation of parameters:

$W_1=\left|\begin{array}{cc}<br /> 0 & xe^{-x} \\ e^{-x}\ln(x) & e^{-x}-xe^{-x}\end{array}\right|=xe^{-2x}\ln(x)$

$W_2=\left|\begin{array}{cc}<br /> e^{-x} & 0 \\ -e^{-x} & e^{-x}\ln(x)\end{array}\right|=e^{-2x}\ln(x)$

$W=\left|\begin{array}{cc}<br /> e^{-x} & xe^{-x} \\ -e^{-x} & e^{-x}-xe^{-x}\end{array}\right|=e^{-2x}$

Thus,

$u_1^/=\frac{W_1}{W}=x\ln(x)$

$\therefore u_1=\int x\ln(x)\,dx$

Apply parts and you get $u_1=\frac{x^2}{2}\ln(x)-\frac{1}{4}x^2$

$u_2^/=\frac{W_2}{W}=\ln(x)$

$\therefore u_2=\int \ln(x)\,dx=x\ln(x)-x$

Therefore, our solution is $y=\left[\frac{x^2}{2}\ln(x)-\frac{1}{4}x^2\right]e^{-x}+\left[x\ln(x)-x\right]xe^{-x}=\color{red}\boxed{\left[\frac{3x^2}{2}\ln(x)-\frac{5}{4}x^2\right]e^{-x}}$
June 15th 2008, 06:06 PM
Mathstud28

Quote:

Originally Posted by Chris L T521

Find the complementary solution:

$y''+2y'+y=0\implies r^2+2r+1=0\implies r=-1$ with multiplicity 2.

Thus, $y_c=c_1e^{-x}+c_2xe^{-x}$

Apply the technique of variation of parameters:

$W_1=\left|\begin{array}{cc}<br /> 0 & xe^{-x} \\ e^{-x}\ln(x) & e^{-x}-xe^{-x}\end{array}\right|=xe^{-2x}\ln(x)$

$W_2=\left|\begin{array}{cc}<br /> e^{-x} & 0 \\ -e^{-x} & e^{-x}\ln(x)\end{array}\right|=e^{-2x}\ln(x)$

$W=\left|\begin{array}{cc}<br /> e^{-x} & xe^{-x} \\ -e^{-x} & e^{-x}-xe^{-x}\end{array}\right|=e^{-2x}$

Thus,

$u_1^/=\frac{W_1}{W}=x\ln(x)$

$\therefore u_1=\int x\ln(x)\,dx$

Apply parts and you get $u_1=\frac{x^2}{2}\ln(x)-\frac{1}{4}x^2$

$u_2^/=\frac{W_2}{W}=\ln(x)$

$\therefore u_2=\int \ln(x)\,dx=x\ln(x)-x$

Therefore, our solution is $y=\left[\frac{x^2}{2}\ln(x)-\frac{1}{4}x^2\right]e^{-x}+\left[x\ln(x)-x\right]xe^{-x}=\color{red}\boxed{\left[\frac{3x^2}{2}\ln(x)-\frac{5}{4}x^2\right]e^{-x}}$

I am sorry you are wrong (Crying)...it is actually $u'$ not $u^{/}$

but besides that it looks good:D
June 16th 2008, 11:47 AM
galactus

Here is a decent one. I like problems like this (it's not that bad):

The conical tank in the drawing loses water out of the hole in the bottom.

The cross-sectional area of the hole is 1/4 ft^2. Determine a DE representing

the height of the water h at ant time. Ignore friction and contraction of the

water stream at the hole.

Hint: you can use

$\frac{dh}{dt}=\frac{-A_{0}}{A_{w}}\sqrt{2gh}$ to

represent the height at any time, but use the given data.
June 16th 2008, 01:08 PM
flyingsquirrel

1 Attachment(s)

Quote:

Originally Posted by galactus

Here is a decent one. I like problems like this (it's not that bad):

The conical tank in the drawing loses water out of the hole in the bottom.

The cross-sectional area of the hole is 1/4 ft^2. Determine a DE representing

the height of the water h at ant time. Ignore friction and contraction of the

water stream at the hole.

Hint: you can use

$\frac{dh}{dt}=\frac{-A_{0}}{A_{w}}\sqrt{2gh}$ to

represent the height at any time, but use the given data.

Torricelli's law tells us that the speed of the water which flows out of the tank is $\sqrt{2gh}$ . (the hole is small enough so that this is correct) As there's no "contraction of the water stream at the hole" (conservation of volumetric flow rate)

$-\frac{1}{4}\sqrt{2gh}=A_{\omega}\frac{\mathrm{d}h} {\mathrm{d}t}$

Substituting $A_{\omega}=\pi\left(\frac{8}{20}h\right)^2=\pi\fra c{4}{25}h^2$ gives us

$<br /> \frac{1}{4}\sqrt{2gh}=-\pi\frac{4}{25}h^2\frac{\mathrm{d}h}{\mathrm{d}t}$

which can be rewritten

$\mathrm{d}t=-\pi\frac{16}{25\sqrt{2g}}h^{\frac{3}{2}}\mathrm{d} h$

Integrating :

$t=\pi\frac{16}{25\sqrt{2g}}\frac{2}{5}\left(8^{\fr ac{5}{2}}-h^{\frac{5}{2}}\right) \Longrightarrow h=\left(8^{\frac{5}{2}}-\frac{125\sqrt{2g}}{32\pi}t\right)^{\frac{2}{5}}$

Using the graph, one can see that the function $h(t)$ is linear for $t<60\,\mathrm{ seconds}$ which could be interesting to make an hourglass. (it'd be a large one :D)
June 16th 2008, 02:11 PM
galactus

Cool Flying squirrel. Here is what I done. It's essentially the same as you. I like your graph which shows it.

We have $A_{0}=\frac{1}{4}$ . To find $A_{w}$ we

note that the radius r corresponding to $A_{w}$ satisfies

$\frac{r}{h}=\frac{8}{20}$ .

Therefore, $r=\frac{2h}{5} \;\ and \;\ A_{w}=\frac{4{\pi}h^{2}}{25}$ .

Then $\frac{dh}{dt}=\frac{\frac{-1}{4}}{\frac{4{\pi}h^{2}}{25}}\cdot\sqrt{2gh}=\fra c{-25\sqrt{2gh}}{16{\pi}h^{\frac{3}{2}}}$
June 16th 2008, 05:21 PM
NonCommAlg

this one is pretty! have fun! (using "Taylor series" is not allowed!)

find the general solution of: $y'' + y' + a^2e^{2x}y=0,$ where $0 \neq a \in \mathbb{R}$ is a constant.
June 18th 2008, 11:25 AM
Chris L T521

Quote:

Originally Posted by NonCommAlg

this one is pretty! have fun! (using "Taylor series" is not allowed!)

find the general solution of: $y'' + y' + a^2e^{2x}y=0,$ where $0 \neq a \in \mathbb{R}$ is a constant.

I've been thinking about this one for a couple days now. I'm not ever sure how to start the problem! (Doh)

I'm assuming that there is some interesting trick needed in order to get things started... :D
June 18th 2008, 05:59 PM
NonCommAlg

Quote:

Originally Posted by Chris L T521

I've been thinking about this one for a couple days now. I'm not ever sure how to start the problem! (Doh)

I'm assuming that there is some interesting trick needed in order to get things started... :D

basically the point of this problem is that it's hard to start it! haha ... anyway, here's how to solve the problem:

let $t=e^x$ and $z=e^xy.$ then see that our differential equation, in terms of $t$ and $z,$ becomes: $\frac{d^2 z}{dt^2} + a^2z=0,$

which is very easy to solve.
June 18th 2008, 06:14 PM
Chris L T521

Here's an interesting problem:

A mass $m=1$ is attached to a spring with constant $k=4$ ; there is no dashpot. The mass is released
from rest with $x(0)=3$ . At the instant $t=2\pi$ the mass is struck with a hammer, providing
an impulse $p=8$ . Determine the motion of the mass.

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