`ydx+(y-x)dy=0`

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

`dx+((y-x)/y)dy=0`

`dx=((-(y-x))/y)dy`

`dx/dy=-(1-x/y)` -------(1)

Now, let x=vy

`rArrdx/dy=v+y(dv)/dy`

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

`v+y(dv)/dy=-(1-(vy)/y)`

`v+y(dv)/dy=-(1-v)`

`y(dv)/dy=-1+v-v`

`y(dv)/dy=-1`

`dv=-dy/y`

on integrating both sides we get,

`v=-log|y|+C` where C is a constant

substitute back the...

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`ydx+(y-x)dy=0`

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

`dx+((y-x)/y)dy=0`

`dx=((-(y-x))/y)dy`

`dx/dy=-(1-x/y)` -------(1)

Now, let x=vy

`rArrdx/dy=v+y(dv)/dy`

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

`v+y(dv)/dy=-(1-(vy)/y)`

`v+y(dv)/dy=-(1-v)`

`y(dv)/dy=-1+v-v`

`y(dv)/dy=-1`

`dv=-dy/y`

on integrating both sides we get,

`v=-log|y|+C` where C is a constant

substitute back the value of v in above,

`x/y=-log|y|+C`

Hello!

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.