Find the general solution to the first order differential equation: ydx + (y-x)dy = 0

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This is a homogeneous differential equation, which can be written as:



`dx/dy=-(1-x/y)`    -------(1)

Now, let x=vy


substitute the values of x and dx/dy in (1)






on integrating both sides we get,

`v=-log|y|+C`   where C is a constant

substitute back the value of v in above,


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Let's pick up `y` as the independent variable. Divide the equation by `dy` and obtain

`y*(dx)/(dy) +y-x=0,`  or  `y*x'(y)-x(y)+y=0.`

Now divide it by `y^2:`    `(y*x'(y)-x(y))/(y^2)+1/y=0.`

Observe that `(y*x'(y)-x(y))/(y^2)=((x(y))/y)'`  and obtain  `((x(y))/y)'=-1/y.`

Now both sides are integrable:

`(x(y))/y=-ln|y|+C,`  or  `x(y)=-yln|y|+Cy,` where `C` is an arbitrary constant.

This is the answer in terms of a function of `y.` There is no way to express `y` as a function of `x` using only elementary functions.

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