1. **State the problem:** Solve the differential equation $$t^2 \frac{dx}{dt} + 2tx - 3t^2 = 0$$ for $x$ as a function of $t$. 2. **Rewrite the equation:** Divide both sides by $t^2$ (assuming $t \neq 0$) to get $$\frac{dx}{dt} + \frac{2}{t}x = 3.$$ This is a linear first-order differential equation. 3. **Identify the integrating factor:** The integrating factor $\mu(t)$ is given by $$\mu(t) = e^{\int \frac{2}{t} dt} = e^{2 \ln |t|} = t^2.$$ 4. **Multiply the entire equation by the integrating factor:** $$t^2 \frac{dx}{dt} + 2t x = 3 t^2.$$ Notice this matches the original equation, confirming the integrating factor. 5. **Rewrite the left side as a derivative:** $$\frac{d}{dt} (t^2 x) = 3 t^2.$$ 6. **Integrate both sides with respect to $t$:** $$t^2 x = \int 3 t^2 dt = t^3 + C,$$ where $C$ is the constant of integration. 7. **Solve for $x$:** $$x = \frac{t^3 + C}{t^2} = t + \frac{C}{t}.$$ 8. **Express the implicit solution:** Multiply both sides by $t^2$ to get $$t^2 x - t^3 = C,$$ which can be written as $$h(t,x) = t^2 x - t^3 = C.$$ **Final answer:** The general solution to the differential equation is $$x(t) = t + \frac{C}{t}$$ or implicitly $$t^2 x - t^3 = C.$$