Suppose that f(x, y) and g(x, y) are homogene | vedclass

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(D) Given the differential equation $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$.Since $f$ and $g$ are homogeneous functions of the same order,we can wri

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$\left(\frac{g(V, 1)}{f(V, 1)}-V\right)$

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(D) Given the differential equation $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$.Since $f$ and $g$ are homogeneous functions of the same order,we can write $\frac{dy}{dx} = \frac{f(Vy, y)}{g(Vy, y)} = \frac{y^n f(V, 1)}{y^n g(V, 1)} = \frac{f(V, 1)}{g(V, 1)}$.We are given the substitution $x = Vy$. Differentiating with respect to $y$,we get $\frac{dx}{dy} = V + y\frac{dV}{dy}$.Thus,$\frac{dV}{dy} = \frac{1}{y} \left( \frac{dx}{dy} - V \right)$.Since $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$,we have $\frac{dx}{dy} = \frac{g(x, y)}{f(x, y)} = \frac{g(Vy, y)}{f(Vy, y)} = \frac{g(V, 1)}{f(V, 1)}$.Substituting this into the expression for $\frac{dV}{dy}$,we get $\frac{dV}{dy} = \frac{1}{y} \left( \frac{g(V, 1)}{f(V, 1)} - V \right)$.Comparing this with $\frac{dV}{dy} = \frac{1}{y}(F(V))$,we find $F(V) = \left( \frac{g(V, 1)}{f(V, 1)} - V \right)$.

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

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