1. **State the problem:** Solve the differential equation $$y^4 y' = 7x^3 + 9x^2 + x + 8\sqrt{x} + 2.$$\n\n2. **Rewrite the equation:** Recall that $y' = \frac{dy}{dx}$, so the equation is $$y^4 \frac{dy}{dx} = 7x^3 + 9x^2 + x + 8x^{1/2} + 2.$$\n\n3. **Separate variables:** We want to write it as $$y^4 dy = (7x^3 + 9x^2 + x + 8x^{1/2} + 2) dx.$$\n\n4. **Integrate both sides:** Integrate with respect to their variables: $$\int y^4 dy = \int (7x^3 + 9x^2 + x + 8x^{1/2} + 2) dx.$$\n\n5. **Compute the integrals:**\n- Left side: $$\int y^4 dy = \frac{y^5}{5} + C_1.$$\n- Right side: Integrate term by term:\n$$\int 7x^3 dx = \frac{7x^4}{4}, \quad \int 9x^2 dx = 3x^3, \quad \int x dx = \frac{x^2}{2},$$\n$$\int 8x^{1/2} dx = 8 \cdot \frac{2}{3} x^{3/2} = \frac{16}{3} x^{3/2}, \quad \int 2 dx = 2x.$$\n\n6. **Write the integrated form:** $$\frac{y^5}{5} = \frac{7x^4}{4} + 3x^3 + \frac{x^2}{2} + \frac{16}{3} x^{3/2} + 2x + C,$$ where $C$ is the constant of integration (absorbing $C_1$).\n\n7. **Solve for $y$ if desired:** Multiply both sides by 5 and take the fifth root: $$y = \left( \frac{35x^4}{4} + 15x^3 + \frac{5x^2}{2} + \frac{80}{3} x^{3/2} + 10x + 5C \right)^{\frac{1}{5}}.$$\n\n**Final answer:** $$y = \left( \frac{35x^4}{4} + 15x^3 + \frac{5x^2}{2} + \frac{80}{3} x^{3/2} + 10x + 5C \right)^{\frac{1}{5}}.$$