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yoyoatw
 one year ago
Find the general solution: y'=(y/x) + [(x^4)e^(y/x)^5]/[5y^4]
yoyoatw
 one year ago
Find the general solution: y'=(y/x) + [(x^4)e^(y/x)^5]/[5y^4]

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yoyoatw
 one year ago
Best ResponseYou've already chosen the best response.1\[y'= \frac{ y }{ x } + \frac{ x^4 * e ^{(y/x)^5} }{ 5y^4 }\]

hink
 one year ago
Best ResponseYou've already chosen the best response.0When you say find the general solution, do you mean solve for the original function of x?

yoyoatw
 one year ago
Best ResponseYou've already chosen the best response.1Basically 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.

hink
 one year ago
Best ResponseYou've already chosen the best response.0I can't get it, I wish you luck with this problem!..

raffle_snaffle
 one year ago
Best ResponseYou've already chosen the best response.0This might help a bit. http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

b87lar
 one year ago
Best ResponseYou've already chosen the best response.0One solution is:\[y(x) = x {\log(5 (const(\log(x))/5))}^{1/5}\]

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @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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