The solution of y{e^{ - x/y}}dx - (x{e^{ - x/ | vedclass

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(A) Given differential equation is $y{e^{ - x/y}}dx - (x{e^{ - x/y}} + {y^3})dy = 0$.Rearranging the terms,we get $y{e^{ - x/y}}dx - x{e^{ - x/y}}dy =

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$\frac{{{y^2}}}{2} + {e^{ - x/y}} = k$

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(A) Given differential equation is $y{e^{ - x/y}}dx - (x{e^{ - x/y}} + {y^3})dy = 0$.Rearranging the terms,we get $y{e^{ - x/y}}dx - x{e^{ - x/y}}dy = {y^3}dy$.Factoring out ${e^{ - x/y}}$,we have ${e^{ - x/y}}(ydx - xdy) = {y^3}dy$.Dividing both sides by ${y^2}$,we get ${e^{ - x/y}} \frac{ydx - xdy}{{y^2}} = ydy$.Recognizing the differential form $d(\frac{x}{y}) = \frac{ydx - xdy}{{y^2}}$,the equation becomes ${e^{ - x/y}} d(\frac{x}{y}) = ydy$.Integrating both sides,let $u = -x/y$,then $du = -d(x/y)$. The integral becomes $\int -{e^u} du = \int y dy$.This results in $-{e^u} = \frac{{{y^2}}}{2} + C$,where $C$ is the constant of integration.Substituting $u = -x/y$ back,we get $-{e^{ - x/y}} = \frac{{{y^2}}}{2} + C$,which can be rewritten as $\frac{{{y^2}}}{2} + {e^{ - x/y}} = k$ (where $k = -C$).

The solution of $y{e^{ - x/y}}dx - (x{e^{ - x/y}} + {y^3})dy = 0$ is

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