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

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

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

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

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

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

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

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

.Sam.
 2 years ago
Best ResponseYou've already chosen the best response.2There'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

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

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

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1if 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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i 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

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

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

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