anonymous
  • anonymous
Find the general solution: y'=(y/x) + [(x^4)e^-(y/x)^5]/[5y^4]
Differential Equations
katieb
  • katieb
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anonymous
  • anonymous
\[y'= \frac{ y }{ x } + \frac{ x^4 * e ^{-(y/x)^5} }{ 5y^4 }\]
anonymous
  • anonymous
When you say find the general solution, do you mean solve for the original function of x?
anonymous
  • 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.

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anonymous
  • anonymous
I can't get it, I wish you luck with this problem!..
raffle_snaffle
  • raffle_snaffle
This might help a bit. http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
b87lar
  • b87lar
One solution is:\[y(x) = x {\log(-5 (const-(\log(x))/5))}^{1/5}\]
anonymous
  • anonymous
\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles Let \(y=vx\), so \(y'=v+xv'\): \[\begin{align*}y'=\frac{y}{x}+\frac{x^4}{5y^4\exp{\left(\dfrac{y}{x}\right)^5}}~~\Rightarrow~~v+xv'&=\frac{vx}{x}+\frac{x^4}{5(vx)^4\exp{\left(\dfrac{vx}{x}\right)^5}}\\\\\\ v+xv'&=v+\frac{x^4}{5(vx)^4\exp{v^5}}\\\\\\ xv'&=\frac{1}{5v^4\exp{v^5}}\\\\\\ v^4e^{{\large v^5}}~dv&=\frac{1}{5x}~dx\end{align*}\] \(\color{blue}{\text{End of Quote}}\)

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