Question

Suppose that \(f(x,y)\) and \(g(x,y)\) are homogeneous functions of same order. If \(x=Vy\) reduces the equation \(\dfrac{dy}{dx}=\dfrac{f(x,y)}{g(x,y)}\) to the form \(\dfrac{dV}{dy}=\dfrac{1}{y}(F(V))\), then \(F(V)=\)

• A. \(\left(\dfrac{f(1,V)}{g(1,V)}-V\right)\)
• B. \(\left(\dfrac{f(V,1)}{g(V,1)}-V\right)\)
• C. \(\left(\dfrac{g(1,V)}{f(1,V)}-V\right)\)
• D. \(\left(\dfrac{g(V,1)}{f(V,1)}-V\right)\)

Hint:

For homogeneous differential equations, if the substitution is \(x=Vy\), then differentiate with respect to \(y\): \[ \frac{dx}{dy}=V+y\frac{dV}{dy}. \] Do not confuse it with the substitution \(y=vx\), where differentiation is done with respect to \(x\).

Answer 1

The Correct Option is D

Step 1: Start from the homogeneous derivative relation.
Given dy/dx = f(x,y)/g(x,y) with f,g homogeneous of same degree n. Use Euler's theorem: x ∂f/∂x + y ∂f/∂y = n f, same for g.

Step 2: Substitute x = Vy and compute dx/dy.
dx/dy = V + y dV/dy. But dx/dy = g(Vy,y)/f(Vy,y) = g(V,1)/f(V,1) by homogeneity.

Step 3: Isolate dV/dy.
V + y dV/dy = g(V,1)/f(V,1) → y dV/dy = g(V,1)/f(V,1) – V → dV/dy = (1/y)[g(V,1)/f(V,1) – V].

Step 4: Final Answer:
F(V) = g(V,1)/f(V,1) – V.