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

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

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

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

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

SithsAndGiggles
 2 years 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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