show that dy/dx = y(x-y)/x+y , if x= y ln (x-y)?

... x = y. ln (x-y) ........................................... (1)

∴ diff... w.r.t. x, by Chain Rule,

... 1 = y. [1/(x-y)].( 1 - y' ) + ln (x-y) · y'

∴ 1 = [y/(x-y)] - [y/(x-y)].y' + ( x/y ).y' ............ from (1)

∴ 1 - [y/(x-y)] = [ (x/y) - ( y/(x-y) ) ].y'

Now solve for y' and simplify.

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I get a different result.

Differentiating (implicitly)

x/y=ln(x-y) one gets

(y-xy')/y^2=(1-y')/(x-y).

Cross-multiplying,

(x-y)(y-xy')=y^2(1-y')

=>xy-x^2y'-y^2+xyy'=y^2-y^2y'

=>xy-2y^2=[x^2-xy-y^2]y'

which leads to

y'=dy/dx=y(x-2y)/(x^2-xy-y^2).

So, I think the result is in error.

if x = y ln(x-y)

x/y = ln (x-y)

e^(x/y) = x-y

differentiate

e^(x/y){ 1/y -x/y^2*(dy/dx)} = 1 -dy/dx

(x-y){1/y - (x/y^2)*dy/dx} = 1 - dy/dx

(x-y)/y - x(x-y)/y^2 *( dy/dx) = 1 - dy/dx

dy/dx{1- x(x -y)/y^2} = {1 - (x-y)/y}

dy/dx{(y^2 -x^2 +xy)/y^2} = { (y-x+y)/y}

dy/dx{ y^2 - x^2 +xy} = y(2y - x)

dy/dx = {y(2y-x)}/{y^2-x^2 +xy} ..........Ans

x= y log( x-y)

differentiating w.r.to x

1 = dy/dx* log(x-y) + y/(x-y) [1- dy/dx]

( x- y) = (x-y) dy/dx log(x-y) +y- y dy/dx

= dy/dx[ (x-y) log(x-y) - y] + y

x- y = dy/dx[ xlog(x-y) - y log(x-y) -y] +y

or (x-y) = dy/dx [x^2/y -- x - y] + y

or ( x- 2y ) = dy/dx( x^2 - xy - y^2) /y

or dy/dx = y( x-2y)/x^2 -xy -y^2) ans