Hi,
Anyone can advise how the above equation can be solved?
Thanks.
wee
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...
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}}
$