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- Thread starter suvadip
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- #1

- Jan 26, 2012

- 183

Can we ask where this system came from?

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This is purely a mathematical problem. It was set in a university exam. I dont have any further details about it.Can we ask where this system came from?

- Feb 13, 2012

- 1,704

From the first equation...Solve:

dx/dt +2y+e^{x}=t^{2}

dy/dt-x+xe^{x}=0

Please help for this simultaneous differential equations

$\displaystyle y= \frac{t^{2} - x^{\ '} - e^{x}}{2} \implies y^{\ '}= t - \frac{x^{\ ''}}{2} - \frac{x^{\ '}\ e^{x}}{2}$ (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...

Kind regards

$\chi$ $\sigma$

Last edited:

- Jan 26, 2012

- 183

The reason I ask - if it's from an exam and it says solve, then itThis is purely a mathematical problem. It was set in a university exam. I dont have any further details about it.

- Jan 29, 2012

- 1,151

I look forward to seeing it! That looks to me like a very difficult non-linear equation.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...

Kind regards

$\chi$ $\sigma$

Last edited by a moderator:

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- Mar 5, 2012

- 9,591

I just checked and found that W|A couldn't solve it.I look forward to seeing it! That looks to me like a very difficult non-linear equation.

Perhaps its solution didn't fit in the margin next to $\chi$ $\sigma$.

- Feb 13, 2012

- 1,704

The only chance to solve second order ODE we are arrived...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...

$\displaystyle x^{\ ''} + x^{\ '} e^{x} + 2 x (1-e^{x}) = 2 t$ (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

$\chi$ $\sigma$

- Jan 26, 2012

- 183

First, I don't believe that using $x = - \ln \lambda$ that (1) becomes (2). Second, even if it did, you simply cannot integrate like this! $t$ is not constant but the independent variable.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

$\chi$ $\sigma$

Last edited:

- Feb 13, 2012

- 1,704

First I'll try to describe 'step by step' how (1) becomes (2). Starting from...First, I don't believe that using $x = - \ln \lambda$ that (1) becomes (2). Second, even if it did, you simply cannot integrate like this! $t$ is not constant but the independent variable.

$\displaystyle x^{\ ''} + x^{\ '}\ e^{x} + 2\ x\ (1-e^{x})=2\ t$ (1)

... we set...

$\displaystyle x= - \ln \lambda \implies e^{x}= \frac{1}{\lambda} \implies x^{\ '}= - \frac{\lambda^{\ '}}{\lambda} \implies x^{\ ''}= \frac{\lambda^{\ '}}{\lambda^{2}} - \frac{\lambda^{\ ''}}{\lambda} $ (2)

... and from (1) and (2)...

$\displaystyle \frac{\lambda^{\ '}}{\lambda^{2}} - \frac{\lambda^{\ ''}}{\lambda} - \frac{\lambda^{\ '}}{\lambda^{2}} - 2\ \ln \lambda\ (1-\frac{1}{\lambda}) = 2\ t \implies \lambda^{\ ''} + 2\ \{t\ \lambda + \ln \lambda\ (\lambda-1)\}=0$ (3)

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 183

$x = - \ln \lambda$ then $x' = - \dfrac{\lambda'}{\lambda}$ - correct

but

$x'' = - \dfrac{\lambda ''}{\lambda} + \dfrac{\lambda'^2}{\lambda^2}$.