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:
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?
Best Answer
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}