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This is the original ODE: $ y^{1/2}y'''+e^{-x}(y'')^{2+c}-(\frac{xy}{x+1})y'=x $ with c is a positive number. $y(0)=1,y'(0)=0,y''(0)=1$

$1st$ question: If x is large, then $ y^{1/2}y'''$ and $-(\frac{xy}{x+1})y'=x $ are the dominant balance terms, but what the DB when x is small?

$2nd$ question: Also, I try to solve this ODE with c=0 by using matlab, but matlab keep showing "busy" for over 2hours.. And I really need the solution with c=0.

Can anyone help me? Thanks!

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    $\begingroup$ $\frac{xy}{1+x}y'$? $\endgroup$ – Dr. Sonnhard Graubner Oct 21 '14 at 6:35
  • $\begingroup$ Why u know it's xy? $\endgroup$ – Math Boy Oct 21 '14 at 6:39
  • $\begingroup$ i don't know it but i thought a bracket was missing $\endgroup$ – Dr. Sonnhard Graubner Oct 21 '14 at 6:41
  • $\begingroup$ A very strange ODE, indeed! $\endgroup$ – Robert Lewis Oct 21 '14 at 6:58
  • $\begingroup$ As far as numerical solutions are concerned, be careful because the solution is likely to go to $\infty$ at finite values of $x$. For example, with $c=0$ and $y(0)=1$, $y'(0) = y''(0)=0$, you get some very large values of $y$ by $x=8$ or so. $\endgroup$ – Robert Israel Oct 21 '14 at 7:25
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Answer to question (1):

When $x\to 0, e^{-x}\to 1$, the leading terms in the equation are:

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

When $x\to\infty$, $\frac{x}{x+1}\to 1$, so the leading term in the equation are: $$y^{1/2}y'''-yy'=x$$

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  • $\begingroup$ but also this equation is difficult to solve and the solution cantains the Bessel-function $\endgroup$ – Dr. Sonnhard Graubner Oct 21 '14 at 6:58
  • $\begingroup$ Yes,but how to show this? I set $y=C(x-x_e)^{-n}$ and got a mess.. $\endgroup$ – Math Boy Oct 21 '14 at 7:02
  • $\begingroup$ @user147893 You may first set $y^{1/2}(x)=z(x)$ to get rid of square root. $\endgroup$ – mike Oct 21 '14 at 7:07
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    $\begingroup$ How to choose ur answer as best answer? I can't find the button.. $\endgroup$ – Math Boy Oct 23 '14 at 4:16
  • $\begingroup$ Thanks a lot. What you may do is accept the answer you like most by click the check-mark. It is underneath the down-vote button on the left side of your question body. $\endgroup$ – mike Oct 23 '14 at 5:47
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According to Maple, the first four terms of the series solution of the differential equation around $x=0$ with initial condition $y(0)=y_0$, $y'(0) = y_1$, $y''(0)=y_2$, are

$$y \left( x \right) =y_{{0}}+y_{{1}}x+{\frac {y_{{2}}}{2}}{x}^{2}-{ \frac {{{\rm e}^{ \left( 2+c \right) \ln \left( y_{{2}} \right) }}}{6 }{\frac {1}{\sqrt {y_{{0}}}}}}{x}^{3}+{\frac {1}{48\,y_{{2}}} \left( 2 \,{y_{{0}}}^{3/2} \left( {{\rm e}^{ \left( 2+c \right) \ln \left( y_{ {2}} \right) }} \right) ^{2}c+4\,{y_{{0}}}^{3/2} \left( {{\rm e}^{ \left( 2+c \right) \ln \left( y_{{2}} \right) }} \right) ^{2}+2\,{y_ {{0}}}^{3}y_{{1}}y_{{2}}+2\,{{\rm e}^{ \left( 2+c \right) \ln \left( y_{{2}} \right) }}{y_{{0}}}^{2}y_{{2}}+{{\rm e}^{ \left( 2+c \right) \ln \left( y_{{2}} \right) }}y_{{0}}y_{{1}}y_{{2}}+2\,{y_{{0}}}^{2}y_ {{2}} \right) {y_{{0}}}^{-{\frac {5}{2}}}}{x}^{4}+O \left( {x}^{5} \right) $$

Of course for this to work you'll want $y_0 > 0$ and $y_2 > 0$.

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  • $\begingroup$ This is pretty cool, give me some time. $\endgroup$ – Math Boy Oct 21 '14 at 7:05
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    $\begingroup$ Take as long as you want. $\endgroup$ – Robert Israel Oct 21 '14 at 7:05

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