## anonymous 2 years ago Find the general solution: y'=(y/x) + [(x^4)e^-(y/x)^5]/[5y^4]

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1. anonymous

$y'= \frac{ y }{ x } + \frac{ x^4 * e ^{-(y/x)^5} }{ 5y^4 }$

2. anonymous

When you say find the general solution, do you mean solve for the original function of x?

3. anonymous

Basically it's asking for what y is equal to. Doesn't have to have the form "y=" but just needs the y' to go away.

4. anonymous

I can't get it, I wish you luck with this problem!..

5. raffle_snaffle

This might help a bit. http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

6. b87lar

One solution is:$y(x) = x {\log(-5 (const-(\log(x))/5))}^{1/5}$

7. anonymous

$$\color{blue}{\text{Originally Posted by}}$$ @SithsAndGiggles Let $$y=vx$$, so $$y'=v+xv'$$: \begin{align*}y'=\frac{y}{x}+\frac{x^4}{5y^4\exp{\left(\dfrac{y}{x}\right)^5}}~~\Rightarrow~~v+xv'&=\frac{vx}{x}+\frac{x^4}{5(vx)^4\exp{\left(\dfrac{vx}{x}\right)^5}}\\\\\\ v+xv'&=v+\frac{x^4}{5(vx)^4\exp{v^5}}\\\\\\ xv'&=\frac{1}{5v^4\exp{v^5}}\\\\\\ v^4e^{{\large v^5}}~dv&=\frac{1}{5x}~dx\end{align*} $$\color{blue}{\text{End of Quote}}$$