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Math Help - Solve dy/dt = 2y + 1 - (9/16)y^2

  1. #1
    wee
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    Solve dy/dt = 2y + 1 - (9/16)y^2

    Hi,

    Anyone can advise how the above equation can be solved?


    Thanks.
    wee
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  2. #2
    Senior Member Peritus's Avatar
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    This is a simple Riccati equation:

    \begin{gathered}<br />
  y' = 2y + 1 - \frac{9}<br />
{{16}}y^2  \hfill \\<br />
   \Leftrightarrow y' =  - \frac{9}<br />
{{16}}\left( {y + \frac{4}<br />
{9}} \right)\left( {y - 4} \right) \hfill \\ <br />
\end{gathered}

    now it is easy to see that y= 4 and y = -4/9 are particular solutions to this equation, the following substitution allows us to find the general solution:
    <br /> <br />
y = u + y_p

    let's choose y_p=4

    thus:


    \begin{gathered}<br />
  u' = 2\left( {u + 4} \right) + 1 - \frac{9}<br />
{{16}}\left( {u + 4} \right)^2  \hfill \\<br />
   \Leftrightarrow u' =  - \frac{5}<br />
{2}u - \frac{9}<br />
{{16}}u^2  \hfill \\ <br />
\end{gathered}

    which is a Bernoulli equation. The substitution that is needed to solve this Bernoulli equation is:

    z = \frac{1}<br />
{u}

    thus we get:


    z' - \frac{5}<br />
{2}z = \frac{9}<br />
{{16}}

    which is a linear first order ODE, and can be easily solved using the integrating factor method. After you finish just back substitute twice to obtain the final solution...
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    You can treat it like a separable equation too. (It's not so bad.)

    After some simple calculations we have y = \frac{{4 + 4ke^{5t/2} }}<br />
{{ke^{5t/2}  - 9}}, where k (or another letter you want) as usually is the constant.
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  4. #4
    wee
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    Thank you very much.
    I got it now.

    z=-\frac{9}{40}+ke^\frac{5t}{2}=\frac{-9+40ke^\frac{5t}{2}}{40}<br />

    u=\frac{1}{z}= \frac{40}{-9+40ke^\frac{5t}{2}}<br />
\Rightarrow y=u+y_p=\frac{40}{-9+40ke^\frac{5t}{2}}+4=\frac{4+160ke^\frac{5t}{2}}  {-9+40ke^\frac{5t}{2}}<br />
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