# Finding a General Solution

• January 21st 2008, 01:28 PM
Finding a General Solution
Not sure if this is the right place to post this cos its off my module called Applied Maths but seemed more of a calculus topic...

$y^{''} + y^{'} + 2y^2 (1 - y) = 0$

Find the value of the constant a so that $y = (1 + e^{ax})^2$ is a solution of the above equation.

Now i just plugged the y = ... bit into the equation but it ended as

... $= \frac{1}{(1 + e^{ax})^2} (1 - a^2 e^{ax} - \frac{2}{(1 + e^{ax})} + \frac{2a^2 e^{2ax}}{(1 + e^{ax})})$
and figured something had gone wrong somewhere as i cant solve it from there.

Any help appreciated
• January 21st 2008, 01:56 PM
Peritus
it's a lot to write...

$
\begin{gathered}
y'' + y' + 2y^2 (1 - y) = 0 \hfill \\
y_{solution} = \left( {1 + e^{ax} } \right)^2 \hfill \\
\end{gathered}

$

$
\begin{gathered}
y' = 2ae^{ax} \left( {1 + e^{ax} } \right) \hfill \\
y'' = 2a^2 e^{ax} \left( {1 + 2e^{ax} } \right) \hfill \\
\end{gathered}

$

substituting these derivatives into the equation:

$
2a^2 e^{ax} \left( {1 + 2e^{ax} } \right) + 2ae^{ax} \left( {1 + e^{ax} } \right) + 2\left( {1 + e^{ax} } \right)^4 \left[ {1 - \left( {1 + e^{ax} } \right)^2 } \right] = 0

$

$
2a^2 e^{ax} \left( {1 + 2e^{ax} } \right) + 2ae^{ax} \left( {1 + e^{ax} } \right) + 2\left( {1 + e^{ax} } \right)^4 \left[ {1 - \left( {1 + e^{ax} } \right)} \right]\left[ {1 + \left( {1 + e^{ax} } \right)} \right] = 0
$

$
2a^2 e^{ax} \left( {1 + 2e^{ax} } \right) + 2ae^{ax} \left( {1 + e^{ax} } \right) + 2\left( {1 + e^{ax} } \right)^4 \left[ { - e^{ax} } \right]\left[ {2 + e^{ax} } \right] = 0

$

$
2e^{ax} \left[ {a^2 + 2a^2 e^{ax} + a + ae^{ax} - \left( {1 + e^{ax} } \right)^4 \left( {2 + e^{ax} } \right)} \right] = 0
$

now the outside exponent can nulify the equation only if a->inf which is not interesting.

$
a^2 + a + \left( {2a^2 + a} \right)e^{ax} = \left( {1 + e^{ax} } \right)^4 \left( {2 + e^{ax} } \right)

$

now because the following equality is supposed to hold for any value of x (y is a solution) then it must hold in particular for x = 0 thus:

$
\begin{gathered}
a^2 + a + 2a^2 + a = 48 \hfill \\
\hfill \\
3a^2 + 2a - 48 = 0 \hfill \\
\hfill \\
... \hfill \\
\end{gathered}

$
• January 21st 2008, 02:02 PM
Forgot to mention that $y^{''}$ means twice differetiated, etc...
Is it just substituting in y = ... into the first equation?
• January 21st 2008, 02:10 PM
WAIT! its meant to be $y = (1 + e^{ax})^{-1}$ sorry
• January 21st 2008, 02:21 PM
Peritus
Quote:

Originally Posted by Deadstar
WAIT! its meant to be $y = (1 + e^{ax})^{-1}$ sorry

well that's a fatal mistake :) anyway it doesn't matter what's importent is you understand the main idea:

Quote:

because the following equality is supposed to hold for any value of x (y is a solution) then it must hold in particular for x = 0
just plug x=0 in the expression you've got and solve for a.