Hi,

Anyone can advise how the above equation can be solved?

Thanks.

wee

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- Feb 27th 2008, 12:35 AMweeSolve dy/dt = 2y + 1 - (9/16)y^2
Hi,

Anyone can advise how the above equation can be solved?

Thanks.

wee - Feb 27th 2008, 01:50 AMPeritus
This is a simple Riccati equation:

$\displaystyle \begin{gathered}

y' = 2y + 1 - \frac{9}

{{16}}y^2 \hfill \\

\Leftrightarrow y' = - \frac{9}

{{16}}\left( {y + \frac{4}

{9}} \right)\left( {y - 4} \right) \hfill \\

\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:

$\displaystyle

y = u + y_p$

let's choose $\displaystyle y_p=4$

thus:

$\displaystyle \begin{gathered}

u' = 2\left( {u + 4} \right) + 1 - \frac{9}

{{16}}\left( {u + 4} \right)^2 \hfill \\

\Leftrightarrow u' = - \frac{5}

{2}u - \frac{9}

{{16}}u^2 \hfill \\

\end{gathered} $

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

$\displaystyle z = \frac{1}

{u}$

thus we get:

$\displaystyle z' - \frac{5}

{2}z = \frac{9}

{{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... - Feb 27th 2008, 06:42 AMKrizalid
You can treat it like a separable equation too. (It's not so bad.)

After some simple calculations we have $\displaystyle y = \frac{{4 + 4ke^{5t/2} }}

{{ke^{5t/2} - 9}},$ where $\displaystyle k$ (or another letter you want) as usually is the constant. - Feb 27th 2008, 06:21 PMwee
Thank you very much.

I got it now.

$\displaystyle z=-\frac{9}{40}+ke^\frac{5t}{2}=\frac{-9+40ke^\frac{5t}{2}}{40}

$

$\displaystyle u=\frac{1}{z}= \frac{40}{-9+40ke^\frac{5t}{2}}

\Rightarrow y=u+y_p=\frac{40}{-9+40ke^\frac{5t}{2}}+4=\frac{4+160ke^\frac{5t}{2}} {-9+40ke^\frac{5t}{2}}

$