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yoyoatw

  • one year ago

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

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  1. yoyoatw
    • one year ago
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    \[y'= \frac{ y }{ x } + \frac{ x^4 * e ^{-(y/x)^5} }{ 5y^4 }\]

  2. hink
    • one year ago
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    When you say find the general solution, do you mean solve for the original function of x?

  3. yoyoatw
    • one year ago
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    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. hink
    • one year ago
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    I can't get it, I wish you luck with this problem!..

  5. raffle_snaffle
    • one year ago
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    This might help a bit. http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

  6. b87lar
    • one year ago
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    One solution is:\[y(x) = x {\log(-5 (const-(\log(x))/5))}^{1/5}\]

  7. SithsAndGiggles
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @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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