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

• Feb 27th 2008, 01:35 AM
wee
Solve dy/dt = 2y + 1 - (9/16)y^2
Hi,

Anyone can advise how the above equation can be solved?

Thanks.
wee
• Feb 27th 2008, 02:50 AM
Peritus
This is a simple Riccati equation:

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

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:
$

y = u + y_p$

let's choose $y_p=4$

thus:

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

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

$z = \frac{1}
{u}$

thus we get:

$z' - \frac{5}
{2}z = \frac{9}
{{16}}$

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, 07:42 AM
Krizalid
You can treat it like a separable equation too. (It's not so bad.)

After some simple calculations we have $y = \frac{{4 + 4ke^{5t/2} }}
{{ke^{5t/2} - 9}},$
where $k$ (or another letter you want) as usually is the constant.
• Feb 27th 2008, 07:21 PM
wee
Thank you very much.
I got it now.

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

$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}}
$