1. The problem is to analyze the differential equation $$t^2 \frac{dx}{dt} + 2tx - 4t^3 = 0$$ and find correct statements about it. 2. First, rewrite the equation in standard form by dividing through by $t^2$ (assuming $t \neq 0$): $$\frac{dx}{dt} + \frac{2}{t}x = 4t$$ 3. This is a first-order linear differential equation of the form: $$\frac{dx}{dt} + P(t)x = Q(t)$$ where $P(t) = \frac{2}{t}$ and $Q(t) = 4t$. 4. To solve, find the integrating factor $\mu(t)$: $$\mu(t) = e^{\int P(t) dt} = e^{\int \frac{2}{t} dt} = e^{2 \ln |t|} = t^2$$ 5. Multiply both sides of the standard form by $\mu(t) = t^2$: $$t^2 \frac{dx}{dt} + 2tx = 4t^3$$ which matches the original equation, confirming the integrating factor. 6. The left side is the derivative of $t^2 x$: $$\frac{d}{dt}(t^2 x) = 4t^3$$ 7. Integrate both sides with respect to $t$: $$t^2 x = \int 4t^3 dt = t^4 + C$$ 8. Solve for $x$: $$x = \frac{t^4 + C}{t^2} = t^2 + \frac{C}{t^2}$$ 9. Important notes: - The solution is valid for $t \neq 0$. - The equation is linear and can be solved using integrating factors. - The general solution contains an arbitrary constant $C$. Final answer: $$x(t) = t^2 + \frac{C}{t^2}$$