Confusing question in Differential first order

[mepeednas]

Q)
I have a first order ODE of the form
dy/dx = F(ax+by+c)/(Ax+By+C) ---> (a,b,c,A,B,C all non zero constants)

Under what condition, does there exist a linear substitution that reduces the equation to one in which the variables are separable?

(A) Never
(B) if aB = bA
(C) if bC = cB
(D) if cA = aC

Ans: I feel it is (B). But there are unclear doubts. My attempt goes below

if a/A != b/B
------------
I am aware that if a/A != b/B, then I can convert this non-homogeneous eqn into homogeneous
1) by eliminating the constants c and C using ah+bk+c=0 and Ah+Bk+C=0 and substituting x with X+h and y = Y+k. Then eqn reduces to the form (aX+bY)/(AX+BY).
2) further by substituting V=Y/X, I can convert the given eqn into a variable separable one.

if a/A = b/B
-----------
But if a/A = b/B(= t say) still I can write it as dy/dx = [t(Ax+By) + c]/(Ax+By+C) and then substituting U for Ax+By (a linear substitution), it again becomes of the form
(1/b)(dU/dx - A)= (tU+c)/U+C
which again is variable separable.

This is the doubt,
-----------------
I have done linear substitution in both the cases (x = X+h and y = Y+k) and have obtained the eqn in variable separable form. So aB = bA is not the only condition, it may as well be aB!=bA. Can anyone clarify?

 

[lippyka]

Answer: The correct answer is (B). In order for a linear substitution to reduce the ODE to a separable form, the condition aB = bA must be satisfied. If a/A ≠ b/B, then it is not possible to make a linear substitution that will reduce the equation to one with separable variables.