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.Sam. Group TitleBest ResponseYou've already chosen the best response.2
e^(x/y) = xy e^(xy^(1)) = xy differentiate, \[[\frac{x}{y^2}\frac{dy}{dx}+\frac{1}{y}]e^{xy^{1}}=1\frac{dy}{dx}\] \[\frac{dy}{dx}(1\frac{xe^{xy^{1}}}{y^2})=1\frac{e^{xy^{1}}}{y}\] \[\huge \frac{dy}{dx}=\frac{1\frac{e^{xy^{1}}}{y}}{(1\frac{xe^{xy^{1}}}{y^2})}\]
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
multiply y^2 to top and bottom and simplify
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[e^{\frac{x}{y}}\times \frac{yxy'}{y^2}=1y'\] is a start
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
or use .sam. method either way
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
\[\text{Final result}\] \[\frac{dy}{dx}=\frac{y \left(e^{x/y}y\right)}{x e^{x/y}y^2}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
@calyne is it clear that the right hand side becomes \[1y'\]?
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
yeah thanks guys
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
wait sam how did you get from \[[−xy2dydx+1y]exy−1=1−dydx dydx(1−xexy−1y2)=1−exy−1y\]
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
wait flutter i mean how did you get from the first equation to the second in your first post
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
There's 2 steps there, 1)multiply e^{xy^{1}} into [−xy2dydx+1y] 2)moving the dy/dx from RHS to LHS then factor dy/dx
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
\[[\frac{x}{y^2}\frac{dy}{dx}+\frac{1}{y}]e^{xy^{1}}=1\frac{dy}{dx}\] \[\frac{xe^{xy^{1}}}{y^2}\frac{dy}{dx}+\frac{e^{xy^{1}}}{y}+\frac{dy}{dx}=1\] \[\frac{dy}{dx}(1\frac{xe^{xy^{1}}}{y^2})=1\frac{e^{xy^{1}}}{y}\] \[\huge \frac{dy}{dx}=\frac{1\frac{e^{xy^{1}}}{y}}{(1\frac{xe^{xy^{1}}}{y^2})}\]
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
can you show me how to equate that to [ y * ( y  e^(x/y) ) ] / [ y^2  xe^(x/y) ] ??
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
i got to go , bump this up for others to solve
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
if we can start with \[e^{\frac{x}{y}}\times \frac{yxy'}{y^2}=1y'\] we can do it step by step, it is essentially algebra from here on in
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i would a) multiply both sides by \(y^2\) b) put everything with a \(y'\) on one side and everything else on the other c) factor \(y'\) out of the terms d) divide
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[e^{\frac{x}{y}}\times \frac{yxy'}{y^2}=1y'\] \[e^{\frac{x}{y}}(yxy')=y^2y^2y'\] \[ye^{\frac{x}{y}}xe^{\frac{x}{y}}y'=y^2y^2y'\] \[y^2y'xe^{\frac{x}{y}}y'=y^2ye^{\frac{x}{y}}\] \[y'(y^2xe^{\frac{x}{y}})=y^2ye^{\frac{x}{y}}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
check my algebra because it is hard to write all this here, but that is the idea
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
last step is to divide and you are done!
 2 years ago
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