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Solve the initial value first order differential equation problem:

$y' = \displaystyle\frac{y^5}{x(1+y^4)},\ y(1) = 1$

\begin{align} \frac{1+y^4}{y^5}dy &= \frac 1x dx\\ \left(\frac 1{y^5} + \frac 1y\right)dy &= \frac 1x dx\\ -\frac 1 {4y^4} + \text{ln}|y| &= \text{ln}|x| + C_1 \end{align}

This is where I get stuck. How do I solve for $y$ at this point?

Wolfram Alpha gave has this following step, which I do not understand at all:


enter image description here


I tried the raising everything as exponents of $e$, but that seems to be a dead end: \begin{align} -\frac 1 {4y^4} + \text{ln}|y| &= \text{ln}|x| + C_1\\ e^{\left(\text{ln} y - \frac 1 {4y^4}\right)} &= e^{\text{ln} x + C_1}\\ y\cdot e^{\left(- \frac 1 {4y^4}\right)} &= x\cdot C_2,\ \text{where $C_2 = e^{C_1}$} \end{align}


P.S. Natural logarithms don't seem to be working: \ln |y| produces $\ln |y|$. I used \text{ln}|y| for $\text{ln}|y|$ instead. Is this a bug?

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    $\begingroup$ Basic answer is that you don't solve for $y$. The solution to the DE will have to be left in implicit form. Happens a lot. $\endgroup$ – André Nicolas Jul 22 '13 at 20:42
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    $\begingroup$ They are making use of the Lambert W function I believe. Just solve for $C_1$ from your general solution. I do not think an elementary solution exists to your ODE. $\endgroup$ – Cameron Williams Jul 22 '13 at 20:42
  • $\begingroup$ @AndréNicolas and Cameron Williams Thank you very much. $\endgroup$ – JDG Jul 22 '13 at 21:06
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You got nearly the conclusion. Let's elevate to power $-4$ after your equation : \begin{align} e^{\ln y - \frac 1 {4y^4}} &= e^{\text{ln} x + C_1}\\ e^{-4\ln y + \frac 1 {y^4}} &= e^{-4\,\text{ln} x - 4\;C_1}\\ \frac 1{y^4}\;e^{\frac 1 {y^4}} &= \frac {C_2}{x^4}\\ \end{align} So that from the definition of the LambertW function $\displaystyle W(z)\;e^{W(z)}=z\,$ we get for $\,z:=\dfrac {C_2}{x^4}$ : $$W\left(\dfrac {C_2}{x^4}\right)=\frac 1{y^4}$$ or $$y^4=\frac 1{W\left(\dfrac {C_2}{x^4}\right)}$$ and the four solutions proposed by Alpha : \begin{align} y&=\frac 1{\sqrt[4]{\left(W\left(\dfrac {C_2}{x^4}\right)\right)}}\\ y&=\frac {-1}{\sqrt[4]{\left(W\left(\dfrac {C_2}{x^4}\right)\right)}}\\ y&=\frac i{\sqrt[4]{\left(W\left(\dfrac {C_2}{x^4}\right)\right)}}\\ y&=\frac {-i}{\sqrt[4]{\left(W\left(\dfrac {C_2}{x^4}\right)\right)}}\\ \end{align}

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  • $\begingroup$ Thank you. I don't understand the Lambert Function. What does the W represent? Is it another constant that I would need to solve for? After reading the Wikipedia entry, I still don't understand how the equations goes from: \begin{align}Y = X e ^ X \; \Longleftrightarrow \; X = W(Y)\end{align} $\endgroup$ – JDG Jul 23 '13 at 3:01
  • $\begingroup$ Could not edit after five minutes: After rereading the Wikipedia, I understand that the W represents any complex number. However, I still do not understand why $Y = X e ^ X \; \Longleftrightarrow \; X = W(Y)$. $\endgroup$ – JDG Jul 23 '13 at 3:08
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    $\begingroup$ @JDG: The Lambert $W$ function is one of these special functions that you shouldn't hope to rewrite in another way. Of course you may expand it in series, rewrite it as iterations and get approximations as detailed in Corless & all 's paper 'On the Lambert W Function' but technically it is defined as the (multivalued) inverse of the function $\;w\mapsto w\;e^w$ so that when you were able to rewrite an equation as $\;y\;e^y=z$ (with $y$ and $z$ any expression) then 'by definition of the $W$ function' this is equivalent to $z=W(y)$. $\endgroup$ – Raymond Manzoni Jul 23 '13 at 9:19
  • $\begingroup$ Okay, I see. Thank you! $\endgroup$ – JDG Jul 23 '13 at 15:55
  • $\begingroup$ Correction : if $\,z\ge -1/e\,$ then $\;y\;e^y=z\,$ becomes $\,y=W(z)$. $\endgroup$ – Raymond Manzoni Feb 3 '18 at 19:01

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