Re: simultaneous differential equation
chisigma said:
From the first equation...$\displaystyle y= \frac{t^{2} - x^{\ '} - e^{x}}{2} \implies y^{\ '}= t - \frac{x^{\ ''}}{2} - \frac{x^{\ '}\ e^{x}}{2}$ (1)
... and inserting (1) into the second equation...
$\displaystyle x^{\ ''} + x^{\ '} e^{x} + 2 x (1-e^{x}) = 2 t$ (2)
... that is a second order ODE the solution of which will be done in a successive post...
The only chance to solve second order ODE we are arrived...
$\displaystyle x^{\ ''} + x^{\ '} e^{x} + 2 x (1-e^{x}) = 2 t$ (1)
... is the elimination of one of the term $x(t)$ or $x^{\ '}(t)$ and that fortunately can be done with the substitution $x (t)= - \ln \lambda (t)$, and doing that we obtain the new ODE...
$\displaystyle \lambda^{\ ''} =- 2\ \{t\ \lambda + \ln \lambda (\lambda -1)\} $ (2)
Using the identity...
$\displaystyle \lambda^{\ ''} = \frac{d \lambda^{\ '}}{d t} = \frac{d \lambda^{\ '}}{d \lambda}\ \frac{d\ \lambda}{d t} = \lambda^{\ '}\ \frac{d \lambda^{\ '}}{d \lambda}$ (3)
... we can write the (2) as...
$\displaystyle \lambda^{\ '}\ \frac{d \lambda^{\ '}}{d \lambda} =- 2\ \{t\ \lambda + \ln \lambda (\lambda -1)\} $ (4)
... and integrating (4) we obtain...
$\displaystyle \frac{\lambda^{\ '\ 2}}{2} = - t\ \lambda^{2} + \frac{1}{2}\ \lambda\ \{\lambda -2\ (\lambda-2)\ \ln \lambda -4\} + c_{1} \implies \lambda^{\ '} = \pm \sqrt {\lambda\ \{\lambda -2\ (\lambda-2)\ \ln \lambda -4\} - 2\ t\ \lambda^{2} + c_{1}}$ (5)
Writing (5) the original problem is 'half solved'... the feasibility of a complete solution will be analized is a successive post...
Kind regards
$\chi$ $\sigma$
