1. We are given the differential equation $$4t^4y^3 \, dt + (3t^4y^2 + 6y^2) \, dy = 0.$$ 2. Rewrite the equation as $$4t^4 y^3 \, dt + (3t^4 y^2 + 6 y^2) \, dy = 0.$$ 3. Divide the equation by $y^2$ (assuming $y \neq 0$): $$4t^4 y \, dt + (3t^4 + 6) \, dy = 0.$$ 4. Express in differential form: $$4t^4 y \, dt = - (3t^4 + 6) \, dy.$$ 5. Separate variables if possible, write as $$\frac{4t^4 y}{3t^4 + 6} \, dt = - \, dy.$$ 6. The equation is separable: $$\frac{4t^4 y}{3t^4 + 6} \, dt + dy = 0.$$ 7. Since the variables mix, check for an integrating factor or exactness. Define $$M = 4t^4 y^3$$ and $$N = 3t^4 y^2 + 6 y^2.$$ 8. Compute partial derivatives: $$\frac{\partial M}{\partial y} = 12 t^4 y^2,$$ $$\frac{\partial N}{\partial t} = 12 t^3 y^2.$$ 9. These are not equal so not exact. Divide entire equation by $y^2$ to simplify: $$4 t^4 y \, dt + (3 t^4 + 6) \, dy = 0.$$ 10. Now check exactness again: $$M = 4 t^4 y,$$ $$N = 3 t^4 + 6.$$ 11. Calculate: $$\frac{\partial M}{\partial y} = 4 t^4,$$ $$\frac{\partial N}{\partial t} = 12 t^3.$$ 12. Not equal again. Try integrating factor depending on $t$ or $y$ only. Assume function of $t$: $$\mu(t),$$ multiplying whole equation. 13. Condition: $$\frac{d}{dy}[\mu(t) M] = \frac{d}{dt}[\mu(t) N].$$ 14. Since $$\mu(t)$$ does not depend on $y$, $$\mu(t) \frac{\partial M}{\partial y} = \mu'(t) N + \mu(t) \frac{\partial N}{\partial t}.$$ 15. Substitute values: $$\mu(t) 4 t^4 = \mu'(t)(3 t^4 + 6) + \mu(t) 12 t^3.$$ 16. Rearranged to differential equation: $$\mu'(t)(3 t^4 + 6) = \mu(t)(4 t^4 - 12 t^3).$$ 17. Divide both sides by $$\mu(t)(3 t^4 + 6)$$: $$\frac{\mu'(t)}{\mu(t)} = \frac{4 t^4 - 12 t^3}{3 t^4 + 6}.$$ 18. Solve differential equation for $$\mu$$: $$\ln|\mu| = \int \frac{4 t^4 - 12 t^3}{3 t^4 + 6} \, dt.$$ 19. Simplify denominator: factor out 3: $$3(t^4 + 2).$$ 20. Write integral: $$\int \frac{4 t^4 - 12 t^3}{3 (t^4 + 2)} \, dt = \frac{1}{3} \int \frac{4 t^4 - 12 t^3}{t^4 + 2} \, dt.$$ 21. Split integral: $$\frac{1}{3} \int \left(4 - \frac{12 t^3}{t^4 + 2}\right) \, dt$$ since $$\frac{4 t^4}{t^4 + 2} = 4 - \frac{8}{t^4 + 2}.$$ Actually, better to do polynomial division or substitution. 22. Perform polynomial division or rewrite numerator: Denote numerator $$4 t^4 - 12 t^3$$ and denominator $$t^4 + 2.$$ Try dividing $$4 t^4 - 12 t^3 = 4(t^4 + 2) - 12 t^3 - 8.$$ 23. Then $$\frac{4 t^4 - 12 t^3}{t^4 + 2} = 4 - \frac{12 t^3 + 8}{t^4 + 2}.$$ 24. Hence the integral becomes: $$\frac{1}{3} \int 4 \, dt - \frac{1}{3} \int \frac{12 t^3 + 8}{t^4 + 2} \, dt = \frac{4}{3} t - \frac{1}{3} \int \frac{12 t^3 + 8}{t^4 + 2} \, dt.$$ 25. For the integral $$\int \frac{12 t^3 + 8}{t^4 + 2} \, dt,$$ note that derivative of denominator is $$4 t^3,$$ related to numerator. Rewrite numerator as $$12 t^3 + 8 = 3 \times 4 t^3 + 8.$$ 26. Substitute $$u = t^4 + 2, du = 4 t^3 dt.$$ 27. Hence $$\int \frac{12 t^3 + 8}{t^4 + 2} dt = \int \frac{3 \, du + 8 \, dt}{u} = 3 \int \frac{du}{u} + 8 \int \frac{dt}{u}.$$ 28. This splits as $$3 \ln|u| + 8 \int \frac{dt}{t^4 + 2}.$$ 29. Thus the integrating factor logarithm is $$\frac{4}{3} t - \frac{1}{3}(3 \ln|t^4 + 2| + 8 \int \frac{dt}{t^4 + 2}) = \frac{4}{3} t - \ln|t^4 + 2| - \frac{8}{3} \int \frac{dt}{t^4 + 2}.$$ 30. The integral $$\int \frac{dt}{t^4 + 2}$$ is non-elementary and typically left as is or expressed via special functions. 31. Therefore the integrating factor $$\mu(t) = e^{\frac{4}{3} t} \frac{1}{t^4 + 2} e^{- \frac{8}{3} \int \frac{dt}{t^4 + 2}}.$$ 32. Multiplying original equation by $$\mu(t),$$ the equation becomes exact. Integration proceeds accordingly but solution is complex due to integral nature. Final note: The equation is not exact but an integrating factor depending on $$t$$ exists with the integral expression above. Further solution requires advanced techniques for $$\int \frac{dt}{t^4 + 2}$$.