How can I find a general solution to the non-linear second order ODE $$ y'' = \frac{x}{y}-1, $$ if there is one expressible in closed form?
So far I have only found the particular solution $y(x)=x$.
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$\begingroup$ We need to know what you have tried and where you get stuck. $\endgroup$– The CountJan 18, 2017 at 18:05
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$\begingroup$ Was there anything more specific in the question? Like the general solution is in the form... $\endgroup$– Teh RodJan 18, 2017 at 18:12
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$\begingroup$ @TheCount Well, in class we mentioned some 6 types of hgher order nonlinear ODEs which we can solve, but I can't fit this one into any of those types, as they all assume that we have y' as well as y''. $\endgroup$– BlazaJan 18, 2017 at 18:19
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$\begingroup$ @TehRod There was nothing more specific. It also wasn't stated that a general solution can be found. $\endgroup$– BlazaJan 18, 2017 at 18:21
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$\begingroup$ This question looks like a pretty hard problem that very unlikely has an explicit solution. $\endgroup$– FabianJan 18, 2017 at 18:24
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2 Answers
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You can consider as two members Emden-Fowler type nonlinear ODE and follow the method in http://www.sciencepubco.com/index.php/ijamr/article/download/723/628
answered Jan 19, 2017 at 12:34
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1$\begingroup$ This seems to give a possible solution, however it's full of typo-like mistakes. I'm having problems understanding how he got the z(1-xz) term in (3.4). I always get just z(1-z) without the x. As far as i can tell, later on he uses that result so it shouldn't be a typo? $\endgroup$– BlazaJan 19, 2017 at 14:14
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HINT
Can we try introducing $ u(x,y) $ differentiating along these lines
$$ y^{\prime2} = x^2 -u(x,y) $$
$$ 2 y{\prime} y{\prime \prime} = 2 x - \left( \frac{ \partial u}{ \partial x} + \frac{\partial u} {\partial y} \frac{dy} {dx} \right) $$
So may be the pde that needs to be solved is
$$ 2y =\frac{\partial u} {\partial x} + \frac{\partial u} {\partial y} \frac{dy} {dx} $$
answered Jan 18, 2017 at 18:42
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$\begingroup$ Unfortunately, I'm taking a course in ordinary DEs, where this problem was asked, so I am not familiar with PDEs and I am not sure what you are getting at. Can you explain a bit more? $\endgroup$– BlazaJan 18, 2017 at 19:09