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

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

Hi,

Anyone can advise how the above equation can be solved?

Thanks.
wee

2. 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...

3. 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.

4. 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}}
$