[Math] differential equation on straight lines

Question

What is the differential equation of straight lines with algebraic sum of the intercepts fixed as k?
answer: $(xy'-y)(y'-1)+ky'=0$
but I need the solution. help me out please

Answer 1

You may know that, among the different ways to write the equation of a straight line, one of them is :

$$\tag{1}\dfrac{x}{x_0}+\dfrac{y}{y_0}=1$$

where $x_0$ is the abscissa of the intercept with the $x$ axis, and $y_0$ is the ordinate of the intercept with the $y$ axis.

(check for example that if $y=y_0$ then $x=0$ ; the same thing when $x=x_0$).

Then, as $x_0+y_0=k$, one may write (1) under the form:

$$\tag{2}\dfrac{x}{k-y_0}+\dfrac{y}{y_0}=1 \ \iff \ y = y_0(1 - \dfrac{x}{k-y_0}) \ \iff \ $$

$$\tag{3}y = \left(\dfrac{y_0}{y_0-k}\right)x+y_0$$

which depends on the single parameter $x_0$.

Now, if, in the left hand side of differential equation

$$\tag{4} \ (xy'-y)(y'-1)+ky'=0$$

we replace $y$ by (3) and $y'$ by $\left(\dfrac{y_0}{y_0-k}\right)$, it is easy to check that we obtain a right hand side in (4) which is $0$ (variable $y_0$ vanishes in the computation).

In a converse way, one could ask whether all solutions of the given differential equation are straight lines with equation (3).