The differential equation \[ \frac{d y}{d x}=\frac{72}{y^{1 / 6}+81 x^{2} y^{1 / 6}} \] has an implicit general solution of the form \( F(x, y)=K \), where \( K \) is an arbitrary constant. In fact, because the differential equation is separable, we can define the solution curve implicitly by a function in the form \[ F(x, y)=G(x)+H(y)=K . \] Find such a solution and then give the related functions requested. \( F(x, y)=G(x)+H(y)= \)

We are given \[ \frac{dy}{dx}=\frac{72}{y^{1/6}+81x^2y^{1/6}}. \] Notice that the common factor in the denominator can be factored out: \[ \frac{dy}{dx}=\frac{72}{y^{1/6}(1+81x^2)}. \] Multiply both sides by \( y^{1/6} \) and by \( dx \) to separate the variables: \[ y^{1/6}\,dy = \frac{72}{1+81x^2}\,dx. \] ### Step 1. Integrate Both Sides **Left Side:** \[ \int y^{1/6}\,dy. \] Recall the power rule for integration: \[ \int y^p\,dy=\frac{y^{p+1}}{p+1}+C. \] Here, \( p=\frac{1}{6} \) so: \[ \int y^{1/6}\,dy=\frac{y^{1/6+1}}{\frac{1}{6}+1}=\frac{y^{7/6}}{\frac{7}{6}}=\frac{6}{7}y^{7/6}. \] **Right Side:** \[ \int \frac{72}{1+81x^2}\,dx. \] Recognize that \( 81x^2=(9x)^2 \). Factor out constants: \[ \int \frac{72}{1+81x^2}\,dx = 72\int \frac{dx}{1+(9x)^2}. \] We know the integral \[ \int \frac{dx}{1+(ax)^2} = \frac{1}{a}\arctan(ax)+C, \] so with \( a=9 \), \[ \int \frac{dx}{1+(9x)^2} = \frac{1}{9}\arctan(9x). \] Thus, \[ \int \frac{72}{1+81x^2}\,dx = 72\cdot\frac{1}{9}\arctan(9x)=8\arctan(9x). \] ### Step 2. Write the Implicit General Solution After integration we have: \[ \frac{6}{7}y^{7/6}=8\arctan(9x)+C. \] This can be rearranged to the form \[ \frac{6}{7}y^{7/6}-8\arctan(9x)=K, \] where \( K \) is an arbitrary constant. ### Step 3. Identify the Functions \( G(x) \) and \( H(y) \) We want the solution in the form \[ F(x,y)=G(x)+H(y)=K. \] A natural choice is \[ G(x)=-8\arctan(9x) \quad \text{and} \quad H(y)=\frac{6}{7}y^{7/6}. \] Thus, the implicit general solution is expressed as \[ \boxed{-8\arctan(9x)+\frac{6}{7}y^{7/6}=K.} \]