1. The problem is to rewrite the differential equation $$\frac{dx}{dt} + t^3 x = \sin(t) x^4$$ after a proper substitution. 2. This is a Bernoulli differential equation of the form $$\frac{dx}{dt} + P(t)x = Q(t)x^n$$ where here, $$P(t) = t^3$$, $$Q(t) = \sin(t)$$, and $$n = 4$$. 3. The substitution for Bernoulli equations is $$v = x^{1-n} = x^{1-4} = x^{-3}$$. 4. Differentiate $$v$$ with respect to $$t$$: $$\frac{dv}{dt} = -3x^{-4} \frac{dx}{dt}$$. 5. Solve for $$\frac{dx}{dt}$$: $$\frac{dx}{dt} = -\frac{1}{3} x^{4} \frac{dv}{dt}$$. 6. Substitute $$\frac{dx}{dt}$$ and $$x^4$$ into the original equation: $$-\frac{1}{3} x^{4} \frac{dv}{dt} + t^3 x = \sin(t) x^4$$. 7. Divide both sides by $$x^4$$ (assuming $$x \neq 0$$): $$-\frac{1}{3} \frac{dv}{dt} + t^3 x^{1-4} = \sin(t)$$. 8. Recall $$v = x^{-3}$$, so $$x^{1-4} = x^{-3} = v$$, thus: $$-\frac{1}{3} \frac{dv}{dt} + t^3 v = \sin(t)$$. 9. Multiply both sides by $$-3$$: $$\frac{dv}{dt} - 3 t^3 v = -3 \sin(t)$$. 10. The transformed linear differential equation is: $$\frac{dv}{dt} - 3 t^3 v = -3 \sin(t)$$. This is the rewritten form after the substitution $$v = x^{-3}$$.