how to obtain this transformation? From the textbook Handbook-of-Ordinary-Differential-Equations_-Exact-Solutions-Methods-and-Problems on page 40 or so, titled

it says

Then later it says (and my question is on this one:)


The question is How did book obtain 1.2.1.3 from 1.2.2.2 when $a_1 x+b_1 y = k(a_2 x+b_2 y)$?
By plugging the above into 1.2.2.2, I could not simplify it to 1.2.1.3.
Any suggestions?
Here are the two pages in the book


 A: The text is a little confuse. Here is what they mean.
There are two cases.

*

*if $a_1x+b_1y\ne k(a_2x+b_2x)$ you can translate $x,y$ in such a way that

$$\frac{a_1x+b_1y+c_1}{a_2x+b_2x+c_2}=\frac{a_1\hat x+b_1\hat y}{a_2\hat x+b_2\hat x}$$
which turns the equation to a homogeneous one. Then you continue the resolution using the general method for homogeneous equations.

*

*if $a_1x+b_1y=k(a_2x+b_2x)$, you have

$$\frac{a_1x+b_1y+c_1}{a_2x+b_2x+c_2}=\frac{k(a_2x+b_2x)+c_1}{a_2x+b_2x+c_2},$$ which is a function of $a_2x+b_2x$. Then you use the general method for functions of $ax+by$.

For homogeneous equations, let $z:=\dfrac yx$. Then
$$y'=(xz)'=z+xz'=f(z)$$ and $$xz'=f(z)-z$$ is separable.

For functions of $ax+by$, let $z:=ax+by$, then
$$z'=a+by'=a+bf(z)$$
is separable.
A: Note that the text says "of the form" or "of the type". This means it refers to the whole class of ODE of this form or type with arbitrary (sufficiently smooth) $f$. In consequence, the $f$ needs not stay the same between the different formulas, the $f$ in (1.2.2.2) can and will in most cases be different from the $f$ in (1.2.1.3).
