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Math Help - DE Marathon

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    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
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    HINT : It's an Exact Equation...
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  3. #3
    Math Engineering Student
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    Quote Originally Posted by Chris L T521 View Post

    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.
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  4. #4
    Eater of Worlds
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    (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.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by galactus View Post
    (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.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    No one is posting any, but here is one I saw before.

    y''+2y'+y=e^{-x}\ln(x)
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    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}}
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    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 ...it is actually u' not u^{/}

    but besides that it looks good
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  9. #9
    Eater of Worlds
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    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.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  10. #10
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by galactus View Post
    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 )
    Attached Thumbnails Attached Thumbnails DE Marathon-tank.png  
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  11. #11
    Eater of Worlds
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    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}}}
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    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.
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  13. #13
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    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!

    I'm assuming that there is some interesting trick needed in order to get things started...
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  14. #14
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    Quote Originally Posted by Chris L T521 View Post
    I've been thinking about this one for a couple days now. I'm not ever sure how to start the problem!

    I'm assuming that there is some interesting trick needed in order to get things started...
    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.
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  15. #15
    Rhymes with Orange Chris L T521's Avatar
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    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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