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Math Help - A differential equation

  1. #1
    Member Jones's Avatar
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    A differential equation

    Hi, i would like some help with this differential equation.

    \begin{cases}<br />
dy/dt +10y = 1 \\<br />
y(1/10) = 2/10<br />
\end{cases}

    This is a separable differential equation but how would you separate the variables in this case?
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  2. #2
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    Quote Originally Posted by Jones View Post
    Hi, i would like some help with this differential equation.

    \begin{cases}<br />
dy/dt +10y = 1 \\<br />
y(1/10) = 2/10<br />
\end{cases}

    This is a separable differential equation but how would you separate the variables in this case?
    \frac{dy}{dt} = 1 - 10y \Rightarrow \frac{dt}{dy} = \frac{1}{1 - 10y}.
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  3. #3
    Member Jones's Avatar
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    Quote Originally Posted by mr fantastic View Post
    \frac{dy}{dt} = 1 - 10y \Rightarrow \frac{dt}{dy} = \frac{1}{1 - 10y}.
    Okay so the next step would be to integrate?

    y = \int \frac{1}{1-10y}
    \rightarrow \frac {1}{10} ln(1-10y) ?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Jones View Post
    Okay so the next step would be to integrate?

    y = \int \frac{1}{1-10y}
    \rightarrow \frac {1}{10} ln(1-10y) ?
    Sloppy notation, try something like:

     <br />
\int\ dt=\int \frac{1}{1-10y}\;dy<br />

    so:

     <br />
t=-\;\frac {1}{10} \ln(1-10y) +C<br />

    Now continue...

    CB
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  5. #5
    Member Jones's Avatar
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    Isn't there an easier way of doing it?

    Maybe you could use the integrating factor e^{\int 10t}
    and multiply both sides of the equation by that?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Jones View Post
    Isn't there an easier way of doing it?

    Maybe you could use the integrating factor e^{\int 10t}
    and multiply both sides of the equation by that?
    Yes an integrating factor will work, but that's not it. But with the correct integrating factor you will get the same answer, it just depends on what you consider an easier method.

    CB
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  7. #7
    Member Jones's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Sloppy notation, try something like:

     <br />
\int\ dt=\int \frac{1}{1-10y}\;dy<br />

    so:

     <br />
t=-\;\frac {1}{10} \ln(1-10y) +C<br />

    Now continue...

    CB
    Right,
     <br />
\frac {1}{10} \ln(1-10y) +C<br />

    Becomes,  e^{1-10y} * \frac{1}{10} \Longrightarrow e^{-10y} *\frac{e}{10}
    and y(1/10) = 2/10 so
    e^{-1} *\frac{e}{10} = \frac{2}{10}

    Hence C must be = \frac{1}{10}
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Jones View Post
    Right,
     <br />
\frac {1}{10} \ln(1-10y) +C<br />

    Becomes,  e^{1-10y} * \frac{1}{10} \Longrightarrow e^{-10y} *\frac{e}{10}
    and y(1/10) = 2/10 so
    e^{-1} *\frac{e}{10} = \frac{2}{10}

    Hence C must be = \frac{1}{10}
    That is confused, you need to work more clearly. I would do this:

    t=-\frac {1}{10} \ln(1-10y) +C

    so:

    -10t=\ln(1-10y)+C

    Now exponentiating both sides:

    c\;e^{-10t}=1-10y

    and so:

    y=\frac{1}{10}[1-c\;e^{-10t}]

    Now we apply the condition y(1/10)=2/10 :

    \frac{2}{10}=\frac{1}{10}[1-c\;e^{-1}]

    so c=-e

    and so:

    y=\frac{1}{10}[1+e^{-10t+1}]

    CB
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  9. #9
    Member Jones's Avatar
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    Hi,

    Thank you. I understand all the algebra. The part that i don't really get is the
    \int \frac {dy}{dt} i think the notation is confusing. What do you end up with after integrating that?
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by Jones View Post
    Hi,

    Thank you. I understand all the algebra. The part that i don't really get is the
    \int \frac {dy}{dt} i think the notation is confusing. What do you end up with after integrating that?
    At no point was this integrated, what was integrated was two different forms of:

    \int \frac {dt}{dy} \;dy

    (see Mr Fantastics post)

    CB
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  11. #11
    Member Jones's Avatar
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    Ok,

    Just one last question.
    If you have the equation y' + 10y = 0

    The solution to this equation is Ce^{-10t}

    What will the general solution be if you have longer expressions
    for example y' +3x^2y-x^2 = 0

    Is it simply e to the power of that polynomial ?
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  12. #12
    Grand Panjandrum
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    Quote Originally Posted by Jones View Post
    Ok,

    Just one last question.
    If you have the equation y' + 10y = 0

    The solution to this equation is Ce^{-10t}

    What will the general solution be if you have longer expressions
    for example y' +3x^2y-x^2 = 0

    Is it simply e to the power of that polynomial ?
    No, as you can verify for yourself by differentiating what you just suggested and substituting into the equation.

    CB
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  13. #13
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    Quote Originally Posted by Jones View Post
    [snip]
    What will the general solution be if you have longer expressions
    for example y' +3x^2y-x^2 = 0

    Is it simply e to the power of that polynomial ?
    The DE can be re-written as \frac{dy}{dx} = x^2 - 3x^2 y = x^2 (1 - 3y) which is readily seperable.

    However, if it was y' + 3y - x^2 = 0, say .....

    This is NOT seperable so now the easiest approach might be to re-write it as y' + 3 y = x^2 and use the integrating factor method.
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