1. **State the problem:** Solve the differential equation $$ (1 + x^2 y^2) y + (xy - 1)^2 x \frac{dy}{dx} = 0 $$ using the substitution $t = xy$. 2. **Rewrite the equation:** Express in terms of $x$, $y$, and $\frac{dy}{dx}$: $$ (1 + x^2 y^2) y + (xy - 1)^2 x \frac{dy}{dx} = 0 $$ 3. **Substitution:** Let $t = xy$. Then, $$ dt = x dy + y dx $$ which implies $$ x dy = dt - y dx $$ 4. **Rewrite $\frac{dy}{dx}$:** $$ \frac{dy}{dx} = \frac{dt - y dx}{x dx} = \frac{dt}{x dx} - \frac{y}{x} $$ 5. **Rewrite the original equation in terms of $t$ and $x$:** Note that $x^2 y^2 = t^2$ and $xy = t$. The equation becomes: $$ (1 + t^2) y + (t - 1)^2 x \frac{dy}{dx} = 0 $$ 6. **Express $y$ in terms of $t$ and $x$:** $$ y = \frac{t}{x} $$ 7. **Substitute $y$ and $\frac{dy}{dx}$:** $$ (1 + t^2) \frac{t}{x} + (t - 1)^2 x \left( \frac{dt}{x dx} - \frac{t}{x} \right) = 0 $$ 8. **Simplify:** $$ \frac{t (1 + t^2)}{x} + (t - 1)^2 \left( \frac{dt}{dx} - t \right) = 0 $$ Multiply both sides by $x$: $$ t (1 + t^2) + (t - 1)^2 x \frac{dt}{dx} - (t - 1)^2 t = 0 $$ 9. **Group terms:** $$ (t - 1)^2 x \frac{dt}{dx} + t (1 + t^2) - t (t - 1)^2 = 0 $$ 10. **Simplify the $t$ terms:** Calculate: $$ t (1 + t^2) - t (t - 1)^2 = t + t^3 - t (t^2 - 2t + 1) = t + t^3 - (t^3 - 2 t^2 + t) = t + t^3 - t^3 + 2 t^2 - t = 2 t^2 $$ 11. **Rewrite the equation:** $$ (t - 1)^2 x \frac{dt}{dx} + 2 t^2 = 0 $$ 12. **Isolate $\frac{dt}{dx}$:** $$ (t - 1)^2 x \frac{dt}{dx} = -2 t^2 $$ $$ \frac{dt}{dx} = - \frac{2 t^2}{x (t - 1)^2} $$ 13. **Separate variables:** $$ \frac{(t - 1)^2}{t^2} dt = - \frac{2}{x} dx $$ 14. **Simplify the left side:** $$ \frac{(t - 1)^2}{t^2} = \left(1 - \frac{1}{t} \right)^2 = 1 - \frac{2}{t} + \frac{1}{t^2} $$ 15. **Integrate both sides:** $$ \int \left(1 - \frac{2}{t} + \frac{1}{t^2} \right) dt = -2 \int \frac{1}{x} dx $$ 16. **Compute integrals:** $$ \int 1 dt = t $$ $$ \int \frac{2}{t} dt = 2 \ln |t| $$ $$ \int \frac{1}{t^2} dt = - \frac{1}{t} $$ $$ \int \frac{1}{x} dx = \ln |x| $$ 17. **Write the integrated form:** $$ t - 2 \ln |t| - \frac{1}{t} = -2 \ln |x| + C $$ 18. **Recall substitution:** $$ t = xy $$ 19. **Final implicit solution:** $$ xy - 2 \ln |xy| - \frac{1}{xy} = -2 \ln |x| + C $$ This is the implicit solution to the original differential equation using the substitution $t = xy$. --- **Summary:** - Used substitution $t = xy$. - Transformed the equation and separated variables. - Integrated both sides. - Expressed the solution implicitly in terms of $x$ and $y$.