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: Use the substitution \(x=Vy\).
Given, \[ x=Vy \] Differentiating with respect to \(y\), \[ \frac{dx}{dy}=V+y\frac{dV}{dy} \]

Step 2: Convert \(\dfrac{dy}{dx}\) into \(\dfrac{dx}{dy}\).
Given, \[ \frac{dy}{dx}=\frac{f(x,y)}{g(x,y)} \] Taking reciprocal, \[ \frac{dx}{dy}=\frac{g(x,y)}{f(x,y)} \] Using \[ x=Vy, \] we get \[ \frac{dx}{dy}=\frac{g(Vy,y)}{f(Vy,y)} \]

Step 3: Use homogeneity of \(f\) and \(g\).
Since \(f(x,y)\) and \(g(x,y)\) are homogeneous functions of the same order, suppose the order is \(n\).
Then, \[ f(Vy,y)=y^n f(V,1) \] and \[ g(Vy,y)=y^n g(V,1) \] Therefore, \[ \frac{g(Vy,y)}{f(Vy,y)} = \frac{y^n g(V,1)}{y^n f(V,1)} \] \[ = \frac{g(V,1)}{f(V,1)} \]

Step 4: Substitute in the differentiated equation.
We have \[ \frac{dx}{dy}=V+y\frac{dV}{dy} \] Also, \[ \frac{dx}{dy}=\frac{g(V,1)}{f(V,1)} \] Thus, \[ V+y\frac{dV}{dy}=\frac{g(V,1)}{f(V,1)} \] \[ y\frac{dV}{dy}=\frac{g(V,1)}{f(V,1)}-V \] Hence, \[ \frac{dV}{dy}=\frac{1}{y}\left(\frac{g(V,1)}{f(V,1)}-V\right) \]

Step 5: Compare with the given form.
Given, \[ \frac{dV}{dy}=\frac{1}{y}(F(V)) \] Therefore, \[ F(V)=\frac{g(V,1)}{f(V,1)}-V \]

Step 6: Final conclusion.
Hence, \[ \boxed{\frac{g(V,1)}{f(V,1)}-V} \]