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Hi all I have a question Ive been asked to solve. But I have no idea where to begin.

The equation is $y'=\dfrac{y+e^x}{x+e^y}$.

I think this is homogeneous but I have no idea as to how to manipulate this to get it into the required form.

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Maybe you can use the fact that you have an equation where $$y' = f\left( {x,y} \right) = \frac{1}{{f\left( {y,x} \right)}}$$ –  Pedro Tamaroff Apr 20 '12 at 18:56
    
I do not think it it homogeneous. –  Fabian Apr 20 '12 at 19:04
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A solution is given by $y(x)=x$ . –  Fabian Apr 20 '12 at 19:14
    
In general, if you have a solution $y=f(x)$ the inverse function $y=f^{-1}(x)$ is also a solution. –  Fabian Apr 20 '12 at 19:50
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Even if the diff-equation cannot be solved, you can get a pretty good understanding of the solution by noting that the function is monotonously increasing and approaches $y=x$ for large $x$. –  Fabian Apr 20 '12 at 19:55
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2 Answers 2

up vote 3 down vote accepted

Maple 16 does not find a closed-form solution, or any symmetries. This strongly suggests that there is no closed-form solution. Almost certainly there are no closed-form solutions that can be found by elementary techniques.

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How about $y(x)=x$ suggested by @Fabian ? –  Sasha Apr 20 '12 at 19:47
    
Neither does Mathematica 8. –  Ayman Hourieh Apr 20 '12 at 19:49
    
Maybe the OP made a typo and forgot a minus sign? –  Fabian Apr 20 '12 at 19:50
    
@Fabian Maybe. If so, it'd be exact and everthing would be OK. –  Pedro Tamaroff Apr 20 '12 at 19:52
    
@Sasha: ok, that's one closed-form solution, but not a general solution. –  Robert Israel Apr 21 '12 at 1:00
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We can write the ode in the form $\omega=Mdx+Ndy=0,$ where $M=-(y+e^x)$ and $N=x+e^y.$
This means that we replace the search for solutions of the ode with the search for curves $\gamma(t)=(x(t),y(t))$ such that $\gamma^\ast\omega=0.$

As showed in Peter Tamaroff's answer $\omega$ is not closed $d\omega\neq 0.$ However it can be showed (invoking Frobenius'theorem) that there exists a function $\mu$ not vanishing s.t. $\mu\omega$ is exact, i.e. $d(\mu\omega)=0.$

To find $\mu$ we need a solution for the $1^{\textrm{st}}$-order linear pde $$0=\frac{\partial \mu N}{\partial x}-\frac{\partial\mu M}{\partial y}\equiv(x+e^y)\partial_x\mu+(y+e^x)\partial_y\mu+2\mu.$$

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... which is as hard to solve as the original ode. –  Robert Israel Apr 20 '12 at 19:47
    
I posted it just to make explicit my difficulty. –  Giuseppe Tortorella Apr 20 '12 at 19:51
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